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# How light falls off

In anything that I’ve read about photography, one is instructed in how light falls off. This is important to know of course. Unfortunately, I also have degrees in physics and electrical engineering, and I already know about light. What the photographers are usually taught is insufficient at best, and dead wrong at worst.

# The inverse square law

The truth that photographers are taught is the inverse square law. I am not trying to say that this is incorrect; it’s just not the whole truth. My guess is that any photographer that ever thought about how they meter scenes would have begun to wonder about its validity as well.

The inverse square law applies to a very specific model. That model involves a point source of light in empty space. To every photographer reading this, I ask you, how many times have you shot this subject?

The essence of the idea comes down to how area works. This will be important in what follows, so it makes some sense to think about this clearly. So, I have a point source of light, and it’s floating in empty space. Say that I am a distance, $D$ away from it. I guess I’m floating in empty space too; neat gig for photography, yes? I set up my camera and meter the light from the point. I get something. Just to be specific, say I get 1/1000s f/8 at ISO 100. Now I float back to twice as far away: $2D$. What do I meter now? The answer is, just like in the photography manuals, 4 times less or 2 stops; for example, 1/1000s f/8 ISO 400.

Here is why. The light from the point source is going out in all directions uniformly. When I am placed at a distance $D$ away from the source, only a small fraction of the total light reaches the area of my lens and falls on my camera’s sensor. Say I want to work out that fraction. I could work out the area of my camera lens. I’d measure it with a ruler, get the radius, $r$ and then figure out $\pi r^2$. The question is, what do I take this area to be a fraction of? The answer is, a sphere of radius $D$.

## The speed of light

Why is that? Well, the answer is thanks to <NAME>, if you must ask. He showed us all how the speed of light is a constant, $c$. Whatever light can reach my camera lens was actually sent from that source some time ago: $D/c$. The other light from the source that was sent out at the same time as what I’m capturing with my camera must, by all the laws of physics, be the same distance away from the source as where I am; that is, $D$. If I go ahead and take a cord of length $D$, hook one end to the light source, and go floating around in my space suit, I’m going to be on some imaginary sphere out in space. Why? Because that’s how you make a sphere: you pick a point and get all the other points a fixed distance away from it in three dimensions.

A point source of light

If you know the formula for the surface area of a sphere (it’s $4/3 \pi D^2$), that’s cool. If you don’t, all you really have to remember is that the area of this sphere will depend on the square of its radius. So, the ratio of the light I can get into my lens to the total light from the point is $\text{Intensity}(\pi r^2)/(4/3 \pi D^2) = 3/4 \text{(Intensity)} r^2/D^2$, for what that’s worth. If you don’t like maths, don’t worry. Just notice that the fraction depends on the square of the distance that I’m from the point. The $\text{Intensity}$ is just some number expressing how intense the light is. It doesn’t matter what this number is; it can be more or less and what I’m working out here is just the same.

If my job as a photographer was capturing images of points of light floating in space, this would be a useful formula to know. You betcha! I make fun, there are some astronomers who work out this kind of thing; but for most of us, this is a rare case.

Anyway, back to the theme… points of light. Now I float back to twice as far, remember? Now the light that hits my lens was sent out actually at a time of $2D/c$ before I hit the shutter button. All of the brother & sister photons that get caught by my camera were sent out at the same time; and they are now on a sphere of radius $2D$. The size of my lens hasn’t changed; I didn’t swap it out. What I got before was

$\text{Light captured at D} = 3/4 \text{(Intensity)} r^2/D^2$

And now I have

$\text{Light captured at 2D} = 3/4 \text{(Intensity)} r^2/(2D)^2$

so the ratio is

$\frac{\text{Light captured at D}}{\text{Light captured at 2D}} = \frac{(2D)^2}{D^2} = 4$

Lot’s of mental anguish for one little number, right? Everything except the square of the ratio of the distances cancels out and, in this case, we get 4. If I’d put some other number in instead of 2, say, $x$, the answer would have been $x^2$.

So this is the inverse square law. If we have a case where the source of light can be modeled as a point, this model makes sense. To be honest, it does make some sense in the context of how light falls off from a source like an unmodified light bulb or flash bulb. So long as we are far enough away from the light source that its physical size is much smaller than the distance, then this formula works.

# A cylinder in space

Now, we move on to another hypothetical situation. Forgive me if this seems artificial. I am heading toward a more practical and universal case, but these stops along the way are important. We have just considered a point source in empty space. Now we consider a long line source. Again, I’m a photographer floating in space some distance $D$ away from a line of light. I’m going to use the same idea as I did with the point source. My camera lens has a certain area. The light from the line source takes a certain time to reach my lens, $D/c$. All the brothers and sisters of the photons that my camera captures left the line at the same time, $D/c$ before. Now, these photons are all on a cylinder of radius $D$. The area of the cylinder is $\pi D L$ where $L$ is the length of the line.

Line of light

We can work out the ratio of the area of my camera’s lens to the area of the cylinder; but the problem is really a lot simpler. All we really need to think about is how the area of the cylinder changes when I move further away or close. That area depends not on the square of the distance, as in the case of the point source, but just on the distance itself. If I move twice as far away, to $2D$ now the area of the cylinder doubles. This is because the area of a cylinder depends directly on the distance, $D$. So the light falls by one stop as I double my distance.

If you want, or can, visualize this, think about the line being comprised of many adjacent points. Each point is sending out its sphere of light, just like in the previous case. Imagine when all of those spheres get to a radius, $D$. What shape do they all make when taken together? The answer is a cylinder.

If you want a practical example, consider the light from a fluorescent tube or one of those linear soft boxes used in studio settings. Photographers sometimes talk about the “quality of light” from different sources. In objective terms, they are talking about how light falls off from the light source, how it shapes itself to the subject. We will see how this works more deeply in the next case we consider.

# A plane of light

Imagine now a plane of light. It stretches out indefinitely in two directions, up/down & left/right, if you will. I am now a photographer floating in space a distance $D$ away from this plane. My lens has its same old area. I am again capturing light from the plane in some ratio that depends on the area occupied by the light that left the plane a time $D/c$ before I hit the shutter button. What is that area? In this case, it is just the area of the plane. What happens if I float back to twice as far, $2D$? Nothing. The area occupied by the light from the plane is just the same. The amount of light captured by my camera doesn’t change at all. Why is this? Because the area of the plane of light at my camera is the same everywhere. It doesn’t drop off at all as I move away from the surface of light.

Think about the plane being comprised of many many points of light. Each one sends out its sphere of light. Imagine this plane coated with these little bubble-like hemispheres of expanding light. Let the spheres overlap each other as densely as the points of the plane. What kind of surface do you get? A plane. Does its area expand as you move away? No. Well, maybe it does at the edges; but in this model, we are floating near the middle of the plane out in space and we can’t even see the edges of it. As far as we are concerned floating there in space, there is no fall off of light at all. If we get the right exposure for the plane, we can float way back or right up; we still have the right exposure.

So while it’s true that the light from any point on the plane falls off as described by the inverse square law, the light from all of the points taken together is a constant. Imagine this light falling on your retina or your camera’s sensor. As you move away, any given patch of light becomes smaller, but light from other patches fills in to compensate exactly. The balance is precise. The intensity of light on your retina or your camera’s sensor remains constant.

## Now we’re getting somewhere

Perhaps you’ve done this yourself. Perhaps you’ve seen it done. A photographer or lighting technician stands in front of a subject and takes a reading with an incident light meter. Perhaps he or she holds the meter right up to the subject’s face. Perhaps he or she stands a bit in front of the subject. The meter measures the total incident light, multiplies by 18%, works out an exposure given some constraints like ISO and aperture, and that’s what the photographer goes with.

That’s what the photographer goes with even if he or she is standing 10 feet or 20 feet back from the subject. Why does this work? If light fell off like the inverse square law, the photographer would have to be adjusting exposures higher by a couple of stops for every time his or her camera distance doubled from where the light meter had been placed. But no one has to do that.

It’s the same idea if one puts an actual 18% gray card in the scene and meters off that. The camera can be moved all over relative to where the card had been placed without any adjustments for proper exposure.

Why is this? Because the scene is an assortment of surfaces. Sure, they have different colors and reflectances and sizes and shapes. However, they behave like that imaginary plane in space. Now, it’s just that the plane has some detail on it. So long as the average is right, the exposure is right.

Of course, there are limits to this idea. If I have some studio lights set up for an outdoor photo shoot at night and I get so far back that my subjects look like a campfire in the distance when I site through my lens, I have exceeded the limits of this model of a plane of light. But that’s so obvious it’s hardly worth mentioning.

On the other hand, if I’m doing landscape photography out in the open where the sun is the source of light, I can move around a scene a lot and not have to readjust exposure. That is, so long as my movements aren’t bringing in some new element with very different lighting characteristics, I can move around a lot. If the average remains right, the exposure remains right.

# An intellectual game

So, suppose I’m out shooting landscapes. I have set an 18% gray card by some beautiful flower. On the other side, I have set a mirror. The sun is behind my back. I have set the mirror up so that it is reflecting the sun directly at my camera. I set my exposure based on the light I’m getting from the gray card. Based on all that I’ve said so far, I should get a beautiful, if surreal, shot, right?

If you agreed with the last statement, we need to do some more work. Let’s think very clearly about this. The sun is a distant source, which means that we have to move quite a distance before its position in the sky changes. That means moving an appreciable part of a time zone, going due east or west. A photographic image that could capture so large a scene would have to be taken from some very high altitude, and that certainly does not apply to my little imaginary scene. So we can assume that the sun is a distant source and all of its light is coming toward the scene at a fixed angle.

## Sunlight

A little research about sunlight tells us that there are roughly 445 Watts per square meter of visible light falling on the surface of the Earth at sea level when the sun is directly overhead. In my intellectual game, I want the sun behind me in the mid-afternoon, call it 3pm. For the sake of argument, imagine that I’ve got my gray card set up so that it is facing the sun at just the right angle. The mirror will be set up at a slightly different angle so that I’m getting the disk of the sun reflected right at me.

Of course, the intensity of the light from the reflection of the sun in the mirror is going to be much much brighter than that from the gray card. Whatever that 18% of maximum from the gray card means, we see right away that the direct light of the sun is more than 100% relative to the gray card; that is, it’s much much greater than $100/18 = 5.555$ times as much. These are the magic 2.47 stops that take us from Value V to the edge of Value VIII.

## Diffuse surfaces scatter

Say that my gray card is big; it’s 1 square meter. I have this big gray card to make the arithmetic easy. So it’s going to scatter back 18% of 445 Watts of visible light. That’s 80 Watts. But what about the mirror? Obviously, it depends on the angle that you look at it. At most viewing angles, you’ll see the sky or the ground or whatever. Only at a very specific angle will you see the sun. This is different from the gray card. You can see its uniform gray surface from any viewing angle at all. That means that as far is the surface of the gray card is concerned, it’s scattering its 80 Watts of light energy into the entire hemisphere that is in front of it.

Also, if I make the gray card bigger or smaller, the amount of light it’s scattering will change in proportion. If it’s 1 square meter, I’ll get 80 Watts into all angles. If it’s 2 square meters, I’ll get 160 Watts, and so on. What about the mirror? I won’t get any more or less sunlight back in that direct ray if I make the mirror larger or smaller, so long as it remains large enough for me to see the full disk of the sun reflected from where I stand.

## Mirrored surfaces reflect

These two surfaces are behaving in very different ways. Let’s try to equalize them. Instead of having a perfect mirror, let’s make an old dingy mirror. In fact, it’s so old and dingy that it reflects only 18% of the light back. Even in this case, the light from the dingy 18% mirror will outshine the 18% gray card. Why?

## Let’s destroy a camera with the sun!

Let’s try an imaginary game that we would never do in practice. We’ll think about the image of the disk of the sun as it is focused on our camera’s sensor. It will have a certain area on the surface of the sensor. In fact, we could cut out a small circle of some given size, paste it onto the sensor, and block out the sun completely. Indeed, there will be some focal length on our lens so that the disk of the sun would just about fill the area of the sensor with its corners dark.

To work this out, we need a mathematical formula or two. First, we want to know what the angle that the disk of the sun falls across at the surface of the Earth. Assuming that it’s 94 million miles away and 900,000 miles in diameter, that comes out to be around 0.55 degrees. Next, we need a formula for the viewing angle of a lens. We can get that from Wikipedia. The formula is

$f = \frac{d}{2 tan(2 \alpha)}$

where $f$ is the focal length, $d$ is the dimension of the sensor, and $\alpha$ is the viewing angle we want. To just fill the vertical dimension of a 35mm camera, we want $d = 24$ mm. We just got the angle $\alpha$ as 0.55 degrees. The answer is that the focal length necessary to just fill the vertical dimension of our 35mm camera is 2508 mm. That’s one whopping lens!

### Kaboom!

Now, we can work out some basic dimensions for the aperture of this lens. At f/1, its diameter is just the focal length, 2508 mm. That means we can work out the area of this whopping lens: it’s going to be $\pi (2.508/2)^2$ in square meters. Ladies and gentlemen of the jury, this means that the area of our fill-the-sensor-with-the-sun lens is almost 5 square meters. With this insane setup, we would be focussing $5 \times 445 = 2225$ Watts of visible light on our poor, and now molten, DSLR sensor!

This isn’t saying anything much that we don’t already know: don’t look right at the sun! It will burn your retina. Likewise, don’t point your camera at the sun: it will burn your sensor. We just quantified the problem.

## Back to sanity

Of course, no one would be insane enough to try this in the real world; and I strongly recommend that you don’t either. Instant destruction of a valuable camera will certainly ensue. But we have a reference point. Suppose we go to a 50mm lens at f/8. Its aperture will be just a circle with a diameter of 6.25 mm. Its area relative to our sensor-killer lens will be $(6.25/2508)^2$ which works out to be about 6.2 parts per million. The total power from the sun goes way down to about 0.014 Watts. Yeah.

This power is focussed onto a much smaller part of the sensor as well because our 50mm lens has a much smaller magnification factor. It doesn’t take too much arithmetic to figure out that the size of the diameter of the disk of the sun on the sensor is now reduced from 24mm to $(80/2508) \times 24 = 0.77$ mm. The power density is now 0.03 Watts per square millimeter. If we knew the number of pixels on the sensor, we could work out the power per pixel. Take my D700 for example. It has 12 million pixels. The area of the sensor is $36 \times 24 = 864$ square millimeters. A pixel has an area of 0.000072 square millimeters. Very small. We get about 2.2 microWatts per pixel. Not so much, right.

What about what we’re getting from our gray card. Recall that it was kicking off 80 Watts into all angles from its surface. Say that we are 3 meters (10 feet) away from it. We have that 50mm lens at f/8. We have a 1 meter object at 3 meters away with a 50mm lens. The real image on our sensor will be about 17 millimeters on a side. (I worked this out from the basic lens formula.) Of the 80 Watts bouncing off this gray card, how much can our lens capture? At 3 meters distance, that 80 Watts is distributed over a hemisphere with an area of $2 \pi 3^2 = 56.5$ square meters. The aperture of my lens at f/8 gives me an area of $\pi (6.25/2)^2 = 30.7$ square millimeters. So, I get a total of $80 \times 30.7 / (56,500,000) = 43.5$ microWatts. This is on an area of $17^2 = 289$ square millimeters. That comes out to cover about 4,000,000 pixels on a D700. On a per pixel basis, we get just $43.5/4,000,000 = 20$ picoWatts per pixel.

## A big big number

Now for the big number! The ratio between the pixels illuminated by the direct reflection of the sun and those illuminated by the gray card. Ready. It’s a factor of $2.2 / 0.00002 = 110,000$! In good old photographic stops, this is about 16 2/3 stops! Again, we’ve just quantified the difference between looking directly at the sun versus looking at scattered light from objects here on Earth, even with a normal lens.

How did this happen? We started out with a perfectly simple setup. We illuminated a gray card in a landscape with sunlight. The sunlight should be just the 100% value relative to the 18% from the gray card, and lo and behold, it is actually more than 16 stops hotter. Not 2.5 stops; 16 stops! When we thought about using a dingy mirror that reduced the light by a factor of 18% (about 2.5 stops), it did almost no good. We’d need a neutral density filter of over 12 stops to get even close, and then the rest of our scene would vanish into blackness.

What the heck just happened? The reason is extremely simple. If we look straight at the sun, we are capturing all of its light. If we look instead at some diffuse element of a scene (like a gray card), we are getting just a tiny fraction of the fraction of the sunlight it has reflected. Why? Because we’re standing back away from it. This time, the inverse square law is working, even though we are still getting the same average amount of light from the entire surface of all of the elements in the scene. This is really the same case as when we were floating in space some distance away from that plane of light. As we back away from the plane, we get less light from the part we could see before (with the fixed viewing angle of our lens) but as that light drops off, we get more light coming in from parts of the plane we couldn’t get into the lens previously. Imagine for a moment that the plane is now a checkerboard pattern of white and black squares. The average is some gray. If I’m far enough away from the plane that my camera captures many squares, I’ll meter the average of the white and black values. If I move away, I get less light from any given white square; obviously, it’s a smaller element on my sensor; but I make up for this by getting the same area of white on my sensor at any position. Exactly half the sensor is white squares no matter how I move. So, my metering value remains the same no matter where I am. (In physics, this is called a symmetry, for whatever that’s worth to you.)

So long as our scene consists only of diffuse surfaces, the maximum intensity of light will be that of a 100% diffuse surface, which is about 2.5 stops above Value V. It is when we have either specular surfaces or direct sources of light that the fun begins. But this already provides us with a useful fact: if we find a surface that spot meters at more than 2.5 stops above Value V, it is a specular surface or a direct light source. It must be.

# Back to Zones

In practical photography, it’s rare to try to capture the direct image of the sun; not impossible, just rare. We can go with a graduated neutral density filter and some HDR techniques and shoot a sunrise; for example. What is much more typical in landscape photography is that a specular surface that isn’t quite a perfect mirror will reflect some significant part of the sun’s direct light back into our camera. In studio photography, we get a similar effect when we catch glare off someone’s eye-glasses or sheen from some metallic object. In event photography, we may capture direct light from stadium lights, street lights, car headlights, whatever.

In every case, the analysis will work out to be very similar to what I just did for a true mirror and the sun relative to a diffuse reference value like an 18% gray card. The ratio may not be quite 16 stops because man-made lighting doesn’t have the intensity of direct sunlight, and real specular surfaces are rarely perfect mirrors, and practical specular surfaces are often not large enough to reflect the entire disk of the sun; but the impact on exposure Values is the same in every case.

That is, it is quite possible to find scenes in which direct light or specular surfaces are much greater than 3 stops above the Value of metered from an 18% diffuse gray card. Heck, it’s not just possible; it’s almost guaranteed.

In landscape work, the classic specular surface is water. Water droplets in clouds, frozen water as snow, the surfaces of streams or lakes, running water on streets, and so on. But there are also metallic surfaces on cars, glossy paint, glass at the right angle, the list goes on. In studio work, similar instances apply.

It becomes essential, then, to accommodate elements in scenes that can significantly exceed Value V; that is, the exposure Value of a properly illuminated 18% diffuse gray card. Unless we’re daft enough to shoot right at the sun, we may not find elements that are 16 stops hotter; but finding elements in scenes that are 5 or 6 stops hotter than middle gray is extremely easy: include a cloud in your shot on a bright day.

We need a strategy for handling this kind of situation, and we still need to get 18% neutral gray right in the final print.

Stay tuned and we’ll see how to manage this.


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# Solving two equations involving Gamma functions where Reduce does not work

In the following lines of code, we are trying to compute the L-moment estimates of the parameters of the Kumaraswamy distribution. But as you notice, neither NSolve nor Reduce work. Interestingly, Mathematica has inbuilt method of moments which very similar to solving the following two equations

 Reduce[
l1 - (β Gamma[1 + 1/α] Gamma[β])/Gamma[1 + 1/α + β] == 0 &&
l2 - l1 + 2 (β Gamma[1 + 1/α] Gamma[2 β])/Gamma[1 + 1/α + 2 β] == 0,
{α, β}]

 Reduce[
0.5296881355 - (β Gamma[1 + 1/α] Gamma[β])/Gamma[1 + 1/α + β] == 0 &&
-0.4161213389 + (2 β Gamma[1 + 1/α] Gamma[2 β])/Gamma[1 + 1/α + 2 β] == 0,
{α, β}]

• Why not use FindRoot? – <NAME> May 8 '18 at 18:10
• @CarlWoll : That is working too! – nutan May 8 '18 at 19:21

This

NMinimize[
Norm[ 0.5296881355-(β Gamma[1+1/α] Gamma[β])/ Gamma[1+1/α+β]]+
Norm[-0.4161213389 + (2 β Gamma[1 + 1/α] Gamma[2 β])/Gamma[1 + 1/α + 2 β]],
{α, β}]

returns this

  {2.22045*^-16, {α->2.4188, β->2.75976}}

in an instant.

That technique can also often provide useful information when there likely is no solution by showing that it found a significantly non-zero minimum

format: plain


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Standard function for Shiny scaffolding.

pivottablerOutput(outputId, width = "100%", height = "100%")

## Arguments

outputId The id of the html element that will contain the htmlwidget. The target width of the htmlwidget. The target height of the htmlwidget.


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zbMATH — the first resource for mathematics

On the orthogonal dimension of orbital sets. (English) Zbl 1077.15025
Let $$V$$ be a complex inner product vector space and let $$e_1,\dots, e_n$$ be an orthonormal basis of $$V$$. Let $$\lambda$$ be a complex irreducible representation of a subgroup $$G$$ of the full symmetric group $$S_m$$. Consider on $$V^{\otimes m}$$ the linear operator $$T(G,\lambda):=\frac{\lambda (\text{id})}{| G| }\sum_{\sigma\in G}\lambda(\sigma)P(\sigma)$$ where $$P(\sigma)$$ is defined by $$P(\sigma)(v_1\otimes\dots\otimes v_m)= v_{\sigma^{-1}(1)}\otimes\dots\otimes v_{\sigma^{-1}(m)}$$.
The authors give a combinatorial necessary and sufficient condition for orthogonality of critical decomposable symmetric tensors $$e^*_\alpha:=T(G,\lambda)(e_{\alpha(1)}\otimes\dots\otimes e_{\alpha(m)})$$ and $$e^*_\beta:=T(G,\lambda)(e_{\beta(1)}\otimes\dots\otimes e_{\beta(m)})$$. The notion of sign-uniform partition is introduced and the set of such partitions is described. The characterization of the sign-uniform partitions is used for obtaining more manageable conditions of orthogonality of $$e^*_\alpha$$ and $$e^*_\beta$$. The concept of orthogonal dimension of a finite set of nonzero vectors is introduced. The orthogonal dimension of critical orbital sets is computed for a class of irreducible characters $$\lambda$$.

MSC:
 15A72 Vector and tensor algebra, theory of invariants
Full Text:
References:
 [1] <NAME>.; <NAME>., Orthogonal decomposable symmetrized tensors, Linear algebra appl., 42, 1-21, (1997) · Zbl 0949.15039 [2] <NAME>., Graphs, (1991), North-Holland Amsterdam · Zbl 0334.05117 [3] <NAME>.; <NAME>.; <NAME>., A note on the orthogonal basis of a certain full symmetry class of tensors, Linear algebra appl., 370, 369-374, (2003) · Zbl 1034.15009 [4] freese, R., Inequalities for generalized matrix functions based on arbitrary characters, Linear algebra appl., 7, 337-345, (1973) · Zbl 0283.15004 [5] <NAME>.; <NAME>., Immanants, Schur functions and the macmahon master theorem, Proc. amer. math. soc., 115, 605-612, (1992) · Zbl 0756.15010 [6] Holmes, R., Orthogonal basis of symmetrized tensor spaces, Linear and multilinear algebra, 39, 241-243, (1995) · Zbl 0831.15018 [7] <NAME>, Orthogonality of cosets relative to irreducible characters of finite groups, preprint · Zbl 1064.20012 [8] <NAME>.; <NAME>., Symmetry classes of tensors associated with certain groups, Linear and multilinear algebra, 32, 21-31, (1992) · Zbl 0762.15015 [9] Littlewood, D.E., The theory of group characters, (1958), Oxford University Press London · Zbl 0079.03604 [10] <NAME>.; <NAME>., Construction of orthonormal bases in higher symmetry classes of tensors, Linear and multilinear algebra, 19, 133-140, (1986) · Zbl 0591.15020 [11] <NAME>., Nonzero decomposable symmetrized tensors, Linear algebra appl., 17, 287-292, (1977) · Zbl 0358.15028 [12] <NAME>., Multilinear algebra, (1997), Gordon and Breach Science Publishers Amsterdam · Zbl 0892.15020 [13] <NAME>.; <NAME>., Inequalities and identities for generalized matrix functions, Linear algebra appl., 64, 223-242, (1985) · Zbl 0563.15006 [14] <NAME>., On the orthogonal basis of the symmetry classes of tensors associated with certain characters, Linear algebra appl., 336, 1-3, 255-260, (2001) · Zbl 0997.15025 [15] <NAME>.; <NAME>., A higher symmetry class of tensors with an orthogonal basis of decomposable symmetrized tensors, Linear and multilinear algebra, 30, 1-2, 61-64, (1991) · Zbl 0735.15022
This reference list is based on information provided by the publisher or from digital mathematics libraries. Its items are heuristically matched to zbMATH identifiers and may contain data conversion errors. It attempts to reflect the references listed in the original paper as accurately as possible without claiming the completeness or perfect precision of the matching.


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# Higgs branching ratios oddity

In reviewing some general articles on theoretical Higgs branching ratios it appears that, other things being equal, more massive decay products are favored over less massive ones. (Bottom quark pair over charm pair, tau pair over muon pair, etc.) This seems unusual since typically phase space factors favor branching to the lighter decay products. Is there a simple, intuitive explanation for this?

• It's not an oddity. It is a cornerstone property of the SM Higgs. Jun 6 '18 at 21:05

## 1 Answer

The coupling of the Higgs boson to a particle is proportional to the mass of that particle. Given the role of the Higgs mechanism in mass generation, this should make some sense intuitively.

The phase space is smaller for more massive daughter particles, but the larger coupling dominates when the combined mass of the daughter particles is not near threshold.

• OK, thanks, I see now that the Higgs-fermion interaction terms after symmetry breaking are proportional to the first power of each fermion mass. Thanks very much, just what I was looking for. Jun 7 '18 at 1:37


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# Eigenvalue with largest imaginary part

Iterative eigensolvers such as ARPACK, give the option to find a subset of the eigenvalues which have the largest imaginary part. My question is how do these algorithms work.

As I understand it, generally such algorithms use Krylov spaces and shift-invert methods which are very good at finding eigenvalues either at the extremes of the spectrum or around a given value. This is essentially because repeated applications of a matrix projects onto the eigenvector with largest magnitude eigenvalue. However, since the imaginary part of an eigenvalue need not have any correlation with its real part, I can't see how it's able to `ignore' the real part.

If anyone has any idea how these algorithms work, or alternatively a way to search for eigenvalues which is ignorant of the real part of the eigenvalue I'd be very grateful.

• You may have more luck here using the numerical-methods tag instead of the computer-science tag. I found it purely by coincidence via scicomp – <NAME> Dec 13 '17 at 23:01


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The conditional expectation is a property associated with a random variable that tells you the likelihood of some event given some other event has already occurred. Before we can discuss this we need to cover some basic definitions first. As always we begin with a probability space $(\Omega,\mathcal{F},\mathbb{P}$ consisting of the sample space of all events and outcomes $\Omega$, the sigma-algebra of possible/measurable events $\mathcal{F}$, and a probability measure that assigns a real number from the unit interval $[0,1]$ to each event $E$ from the sigma-algebra. This is elementary stuff. Next we introduce the conditional probability measure as being a functional of some certain event that has already occurred $A$ (such that $\mathbb{P}(A) = 1$), and another event $E$ whose probability we wish to measure. The first thing to understand is that the probability that some event $E$ will occur can be very much different to the probability that the same event will occur given that some other event has already occurred. This is because the conditional probability measure $\mathbb{P}_A$ (as opposed to the ordinary probability measure $\mathbb{P}$) actually removes the event that has already occurred from the sample space and re-assigns all the probabilities to the remaining to-be-determined events. It is an entirely new measure.

DEFINITION: Conditional Probability Measure
Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space and let $A \subset \mathcal{F}$ be an event that has already occurred with probability one. Then the functional defined by

$\mathbb{P}_A(E)=\mathbb{P}(E|A):=\frac{\mathbb{P}(E \cap A)}{\mathbb{P}(A)}$

is called a conditional probability measure.

#### Partitions

The next ingredient we need is that of a partition, which is simply the act of splitting up a set $\Omega$ in to smaller subsets $\{B_1,\dots,B_n\}$. These smaller subsets are called blocks and they satisfy the property that they never overlap, which means that their intersection is empty: $B_i \cap B_j = \emptyset$; and if you stick all of them together you end up reconstructing the whole set that you began with, which means that their union is the whole set: $B_1 \cup \cdots \cup B_n = \Omega$. We say that one can partition a set in to a collection $\mathcal{P}$ of smaller blocks $\{B_1,\dots,B_n\}$ who are each collectively exhaustive and mutually exclusive with respect to the set being partitioned. There is no mathematical trickery going on here, we are simply breaking up the set, which can be done in a number of different ways, but more importantly, can always be done.

Now why are we interested in partitioning a set? Well, if the set to be partitioned is the sample space $\Omega$ then we should be very interested because if you think about what a random variable does: it maps events (subsets of the sample space) to the real numbers, and now think about the inverse of a random variable: it maps a real number to the sample space (to the sigma-algebra really, but what goes in to the sigma-algebra always goes in to the sample space). Then, thinking about that some more, we can imagine that, depending on the random variable mapping, the same real number may get mapped to the same subsets, so you have various subsets of the sample space with the same number associated to them by the inverse random variable. If you cut up the sample space by these numbers you are partitioning it! Kind of like painting by numbers. Said another way: A random variable naturally partitions the sample space. Let’s see exactly why this is the case.

#### Refinements

Okay, so we are focusing on discrete probability theory here, and in particular discrete random variables. We have at our disposal the idea of partitioning a set, or the sample space, in to a collection $\mathcal{P}$ of blocks. We’ve also hinted that in some way, a random variable partitions the sample space as well. Before we can prove that we need to introduce a very special kind of partition: a refinemnet.

DEFINITION: Refinement
Let $\mathcal{P}$ be a partition of a set $\Omega$. Then a partition $\mathcal{R}=\{B_1^{\prime},\dots,B_n^{\prime}\}$ is called a refinement of $\mathcal{P}$ if each block $B_i^{\prime}$ of $\mathcal{R}$ is contained in some block $B_i$ of $\mathcal{P}$, such that $B_i = \bigcup B_i^{\prime}$. We denote such a refinement this by $\mathcal{P} \prec \mathcal{R}$.

We will need this concept of a refinement when it comes time to prove the connection between random variables naturally inducing a partition on the sample space and it’s measurability.

#### Random Variables and Natural Partitions

Consider a (discrete) random variable $X$, recall that it maps events (or subsets of the sample space) from the sigma-algebra to the real numbers. Think of all the real numbers $x_i$ that the random variable might map to. Collect these together and form a set $\{x_1,\dots,x_n\}$, this is called the image of the random variable – it is a collection of all the points where the random variable mapping sends events and subsets to. We denote this by $\Im(X) = \{x_1,\dots,x_n\}$. Now, the inverse random variable $X^{-1}$ goes backward and maps a real number $x_i$ to a subset $E$ of the sample space $\Omega$. If we know the random variable then we know its image $\Im(X)$, then all we need to do is follow the real numbers back via the inverse random variable and see what subset of the sample space we end up in. Then, as you would with paint-by-numbers, we partition the sample space using those numbers. That is, we naturally partition $\mathcal{P}_X$ the sample space using the inverse random variable, or by partitioning using the inverse image of the elements from $\Im(X)$. In symbols: $\mathcal{P}_X:=\{X^{-1}(x)\,|\,x\in \Im(X)\}$.

#### Measurable Random Variables

If you begin with a probability space (a sample space, a sigma-algebra of events, and a measure functional) and introduce a random variable as a mapping from the sigma-algebra to the real numbers, how do you say that the random variable is measurable? After all, the measure functional that comes equipped with the probability space measures probabilities of events, not random variables. But, with the help of partitions and refinements, we can measure a random variable. Here’s how we do it:

We take a probability space and any random variable $X$. We then form a partition $\mathcal{P}_X$ of the sample space $\Omega$ in to the one induced naturally by the random variable. Since the blocks of the partition have the exact same real number mapped to them by the inverse random variable, we know for sure that the random variable is constant on each block. In other words, the random variable is constant on each block precisely because we partitioned the blocks using the inverse random variable. If we used some other partition then it would no longer be constant. Now we define it:

DEFINITION: Random Variable Measurability
Let $\mathcal{P}$ be the partition of the sample space $\Omega$. A random variable $X$ is said to be $\mathcal{P}$-measurable if the random variable is constant on each block $B_i$ of the partition $\mathcal{P}$.

Could the same random variable might be considered measurable given some other partition and not the natural one? Maybe. It depends on whether or not the partition is a refinement of the natural partition.

THEOREM
A random variable $X$ is said to be $\mathcal{R}$-measurable if and only if $\mathcal{R}$ is a refinement of the natural partition $\mathcal{P}_X$.

#### What About Infinite Sample Spaces?

We have seen so far that for finite sample spaces, partitions are intimately connected with random variables via their inverse images. As it happens, the notion of a partition does not generalise readily for infinite sample spaces. Instead we use an algebra.

Recall that an algebra $\mathcal{A}$ is a collection of subsets of some larger set $\Omega$ such that the empty set is in the collection, it’s closed under complements, and under unions. Now, just as a random variable can naturally define a partition of the sample space, it can also define a natural algebra $\mathcal{A}_X$ consisting, again, of inverse images. Now let us re-define the measurability of a random variable to one with respect to an algebra instead of a partition. In fact, it comes as no surprise that for a random variable to be algebra-measurable (instead of partition-measurable) we want it to be constant on all elements of the algebra.

DEFINITION: Measurability of Random Variables
Let $X$ be a random variable on a sample space $\Omega$. Let $\mathcal{A}$ be some algebra of subsets on $\Omega$. Then the random variable $X$ is $\mathcal{A}$-measurable if $X$ is equal to some constant $C$ on all elements of the algebra, or:

$\{X=c\}\in\mathcal{A},\,\forall\,c\subseteq \Im(X)$

Continuing in the same fashion as before, we can say that any random variable is measurable with respect to any algebra $\mathcal{A}$ so long as the algebra is (not a refinement) but simply a subset of the algebra naturally induced by the random variable: $\mathcal{A}$

#### Finally, Conditional Expectation!

We are now ready to formulate the definition of conditional expectation: We just take the ordinary expectation of a random variable of an event $E$, but now we do it using the conditional probability measure $\mathbb{P}_A$, where $A$ is some certain event that has already occurred. Thus,

DEFINITION: Conditional Expectation
Let $(\Omega, \mathcal{F},\mathbb{P})$ be a probability space and let $A$ be an event which has or could potentially occur. Then the conditional expectation (using the ordinary probability measure $\mathbb{P}$that comes equipped with the probability space) of a random variable $\mathbb{E}^{\mathbb{P}}[X|A]$ equals the expectation (using the new conditional probability measure $\mathbb{P}_A$) of just the random variable with no conditioning. In symbols:

$\mathbb{E}^{\mathbb{P}}[X|A] = \mathbb{E}^{\mathbb{P}_A}[X]$

If we use the indicator function $\mathbb{I}_A$, which returns 1 when it eats a subset of $A$ and a 0 when it eats something which is not, and we use it on the random variable $X$, we get the product $X\mathbb{I}_A$ which essentially forces the random variable to only act on subsets of the specified subset $A$ (the result otherwise is zero which, in probability terms, is precisely the same as an impossible event). Thus the product $X\mathbb{I}_A$ can be viewed in terms of set operations as $X \cap A$ – this is clear abuse of notation as $X$ is not a set, hence why we need the indicator function to do the same thing as set inclusion. Using the ordinary probability measure $\mathbb{P}$, we can take the ordinary expectation of this $\mathbb{E}^{\mathbb{P}}[X\mathbb{I}_A]$ and then divide this number (remember that the expectation operator $\mathbb{E}$ is a functional which assigns real numbers to events) by the probability of the event $A$ occurring and we get another representation of the conditional expectation that really lines up with the representation of the conditional probability measure introduced at the beginning of this article:

LEMMA: Conditional Expectation
Given a probability space, a random variable $X$ and an event $A$ that has or could occur, then the conditional expectation of $X$ with respect to the event $A$ can be expressed as

$\mathbb{E}^{\mathbb{P}}[X|A] = \frac{\mathbb{E}^{\mathbb{P}_A}[X\mathbb{I}_A]}{\mathbb{P}(A)}$


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Affine Transformation Opencv

OpenCV essentially stands for Open Source Computer Vision Library. An affine transform produces parallelograms (which includes standard rectangles). Lines detection with Hough Transform - OpenCV 3. getAffineTransform(). This is a port of Korman et al implementation of Fast Affine Template Matching in C++. non-uniform scaling in some. (This paper is easy to understand and considered to be best material available on SIFT. import numpy as np import cv2 from matplotlib import pyplot as plt. Note that the matrix form of an affine transformation is a 4-by-4 matrix with the fourth row 0, 0, 0 and 1. Learn to apply different geometric transformation to images like translation, rotation, affine transformation etc. From the Cartesian grid (left grid), we can see the blue point is located at (2, 1). We scale the image by half and rotate it by 90 degrees anticlockwise. i would appreciate it if you could help me with this. affine optic flow. OpenCV provides two transformation functions, cv2. AffineTransform(). Obviously the images represent only partially the same thing (some background is removed and other is added) but after a roto-translation (an affine transformation caused by the camera movement). The purpose of this blog post is to demonstrate how to align a face using OpenCV, Python, and facial landmarks. 4 with python 3 Tutorial 14 by <NAME> February 15, 2018 Beginners Opencv , Tutorials 2. warpPerspective, with which you can have all kinds of transformations. Returns the cross product of *this and other using only the x, y, and z coefficients. An affine transform is a general linear 00059 * transformation from one coordinate space to another. 仿射变换(Affine Transform) 5. getPerspectiveTransform; Transformations. Translation and Euclidean transforms are special cases of the Affine transform. They are extracted from open source Python projects. 4 Point OpenCV getPerspectiveTransform Example. We have been using the function warpAffine quite a bit over the last - Selection from OpenCV with Python By Example [Book]. pixel intensity values located at position in an input image) into new variables (e. GIMP Tutorial – using the Perspective Tool by <NAME> on YouTube. OpenCV 3 is a native cross-platform library for computer vision, machine learning, and image processing. 1、仿射变换（Affine transformation） 因此，空间变换中的仿射变换对应着五种变换，平移，缩放，旋转，翻转，错切。而这五种变化由原图像转变到变换图像的过程，可以用仿射变换矩阵进行描述。. Use the OpenCV function cv::getRotationMatrix2D to obtain a $$2 \times 3$$ rotation matrix; Theory What is an Affine Transformation? A transformation that can be expressed in the form of a matrix multiplication (linear transformation) followed by a vector addition (translation). Raspberry Pi Camera + Python + OpenCV (Day2) 1. Computer vision is a rapidly growing field devoted to analyzing, modifying, and high-level understanding of images. In this recipe, we will review two main ways to geometrically transform images: affine and perspective warps. Converting from x/y back to row/col uses the inverse of the affine transform. The usual way to represent an Affine Transform is by using a 2 \times 3 matrix. Affine Cameras • Example: Weak-perspective projection model • Projection defined by 8 parameters • Parallel lines are projected to parallel lines • The transformation can be written as a direct linear transformation 2x4 projection matrix 2x2 intrinsic parameter matrix 2x3 matrix = first 2 rows of the rotation matrix. Affine transformations In this section, we will discuss about the various generalized geometrical transformations of 2D images. These are all 'linear' distortions, by which I mean two straight parallel lines present in an image will remain straight and parallel. Shows how to do a perspective transform using GIMP. Then I will segue those into a more practical usage of the Python. Note that these 3 sets are actually representing 3 orthogonal axes;. OpenCV format, x is width. Any hints are welcome. Translation and Euclidean transforms are special cases of the Affine transform. Now I'm looking into perspective transformation and Region of Interest but have no success so far. geometric_transform (input, mapping[, …]) Apply an arbitrary geometric transform. Set expected transformation to affine; Look at estimated transformation model [3,3] homography matrix in ImageJ log. Before we jump into the process of face detection, let us learn some basics about working with OpenCV. Geometric Image Transformations¶. 4 Point OpenCV getPerspectiveTransform Example. I concluded by demonstrating how the same can be done using two popular Python libraries Pillow and OpenCV. cvtColor(input_image, flag) where flag determines the type of conversion. The functions in this section perform various geometrical transformations of 2D images. GIMP Tutorial - using the Perspective Tool by <NAME> on YouTube. 기본적으로 제공하는 library가 매우 빵빵하기 때문이다. Image transformation techniques like color spaces, affine and perspective transforms. Search Search. Projective transformations a. Affine Transformation是一种坐标之间的线性变换，保持图形的“平直性”（译注：straightness，即变换后直线还是直线不会打弯，圆弧还是圆弧）和“平行性”（译注：parallelness，其实是指保二维图形间的相对位置关系不变，平行线还是平行线，相交直线的交角不变。. Basically, given a source x,y point, affine matrix for the transformation account for the center of my image actually being width/2 and height/2, and the visual bounds I would even be able to grab from in the non transformed image, then I warp the x and y affined points to the correct undistorted plane and return the calculated row and column i. What kind of transformation is T? How can we compute the transformation matrix: T(x,y )? A B C A' C' B' Source Destination = 1 0 0 1 1 ' ' y x d e f a b c y x Two ways: Algebraic and geometric. Affine Transformation. - Define two functions to implement the process of points selection - Compute the affine transformation - Compute the perspective transformation. Obviously the images represent only partially the same thing (some background is removed and other is added) but after a roto-translation (an affine transformation caused by the camera movement). I want to perform the affine transformation on a set of points. I saw your Tutorial 12 and would like to ask you how i can do the same with cv2. In this video, we'll use OpenCV's built-in functions for warping an image using affine and perspective transformations. Stitching images using template matching and warpAffine (self. 仿射轉換 Affine Transformation 2010年4月14日 星期三 仿射轉換 Affine Transformation; OpenCV Dev C++ 設定. But in some cases, you may need elliptical/circular shaped kernels. The size of *this and other must be four. It has four degrees of freedom and requires two pairs of points. warpAffine(arr, M, (cols, row)) this works if the image is represented as bitmap. 2 Transformations transformations Preservation of affine combinations of points. Detectors evaluation - Matlab files to compute the repeatability. Here is a simple OpenCV C++ example code to rotate an image. So the difference between a $2\times3$ matrix and a $4\times4$ matrix was only from your way of writing it; this works the same way as an affine transform in three dimensions, just with one fewer dimension. Working Subscribe Subscribed Unsubscribe 5K. 4 with python 3 Tutorial 14. You can define the transformations with mouse clicks. interpolation. OpenCV Python – Rotate Image In this tutorial, we shall learn how to rotate an image to 90, 180 and 270 degrees in OpenCV Python with an example. Sets of parallel lines remain parallel after an affine transformation. Affine transformation - OpenCV 3. From the above, We can use an Affine Transformation to express: Rotations (linear transformation) Translations (vector addition) Scale operations (linear transformation) you can see that, in essence, an Affine Transformation represents a relation between two images. The typical geospatial coordinate reference system is defined on a cartesian plane with the 0,0 origin in the bottom left and X and Y increasing as you go up and to the right. Parameters:. Given a set of facial landmarks (the input coordinates) our goal is to warp and transform the image to an output coordinate space. e calculate the transformation matrix required to convert first image into second image. Use the OpenCV function cv::getRotationMatrix2D to obtain a $$2 \times 3$$ rotation matrix; Theory. This function is especially useful when using 4D vectors instead of 3D ones to get advantage of SSE/AltiVec vectorization. import numpy as np import cv2 from matplotlib import pyplot as plt. src Coordinates of triangle vertices in the source image. Once we have the affine transformation matrix, we use the function to apply this matrix to the input image. These steps include subtracting a mean pixel value and scaling the image. To get the transformation matrix, we have a function called in OpenCV. Clone via HTTPS Clone with Git or checkout with SVN using the repository’s web address. It always fun to do affine transformation on image, since some of the results look hilarious. 仿射变换 affine transformation ; 4. 仿射变换(Affine Transformation) 3. ارسال شده در ۱۳۹۶-۱۲-۲۸ ۱۳۹۶-۱۲-۲۸ نویسنده بردیا اسمعیلی دسته‌ها openCV, یادداشت‌های کلاسی برچسب‌ها Affine, opencv, Perspective, transformation, تبدیل, تبدیلات, ماتریس راهبری نوشته. You can vote up the examples you like or vote down the exmaples you don't like. An affine transformation is a 2-dimension cartesian transformation applied to both vector and raster data, which can rotate, shift, scale (even applying different factors on each axis) and skew geometries. Is it possible to fill the missing area from neighboring frames so the that video completion is pleasant. An affine transformation is an important class of linear 2-D geometric transformations which maps variables (e. Section 1 – Image Pre-Processing: In this section, we’ll describe the pre-processing steps that are applied to an input image. The following shows the result of a affine transformation applied to a torus. Multi-dimensional image processing (scipy. float32() array_tform = cv2. 3) Just do the warping on the CPU side using, say, OpenCV or scipy. affine transformation. OpenCV’s estimateRigidTransform is a pretty neat function with many uses. Face alignment with OpenCV and Python. This is the C function called by the opencv C++ function and as a extra benefit, it return false if the affine matrix cannot be solved. Affine transformation. OpenCV provides two transformation functions, cv2. pyplot as plt from skimage. Output Input. Geometric Image Transformations¶ The functions in this section perform various geometrical transformations of 2D images. 4 with python 3 Tutorial 14 by <NAME> February 15, 2018 Beginners Opencv , Tutorials 2. From the image moments , the center mass (which gives the translation), the traffic sign orientation (which gives the rotation) and the principle components (which give the scaling) are computed. OpenCV format, x is width. Homographies "keystone" distortions Finding the transformation How can we find the transformation between these images? Finding the transformation Translation = 2 degrees of freedom Similarity = 4 degrees of freedom Affine = 6 degrees of freedom Homography = 8 degrees of freedom. Python, OpenCVで画像の幾何変換（線形変換・アフィン変換・射影変換）を行うには関数cv2. You can define the transformations with mouse clicks. I want to perform the affine transformation on a set of points. For example, if the affine transformation acts on the plane and if the determinant of is 1 or −1 then the transformation is an equiareal mapping. I personally find it is easier to separate the two, so the view transformation can be modified independently of the model matrix. The result is: Affine Transformation. 1) transform vertices to the coordinate system of the camera, 2) transform the result to the 2D plane of the screen, 3) draw triangles using the resultant 2D vertices. Hence in affine transformation the parallelism of lines is always preserved (as mentioned by EdChum ). ndimage) Apply an affine transformation. GIMP Tutorial – using the Perspective Tool by <NAME> on YouTube. What I would like to do is something like : Mat affineTransformation. U and X are each 3-by-2 and2 and • define the corners of input and output triangles. That is it will modify an image to perform all four of the given distortions all at the same time. Consider the following figure (the length of the sides in the square is 1): The unit square is transformed into the rectangle by using scaling and rotation. Complex geometric transformations (distortion) approximation by partitioning an image into smaller rectangular subimages; for each subimage, a simple geometric transformation, such as the affine, is estimated using pairs of corresponding pixels. OpenCV and Python versions: This example will run on Python 2. By deffault few Opencv provides border extrpolation methods like- REFLECT,REPLICATE,WRAP. Please visit LearnOpenCV. 4+ and OpenCV 2. Sample Code. It comes in a form like: [1. OpenCV入门学习（二）Affine Transformations（仿射变换） 共有140篇相关文章:仿射变换VS透视变换 opencv之仿射变换 使用Gnome Canvas实现图像缩放 空间变换及其应用(二)----仿射,双线性,透视 matlab 中的仿射变换函数 OpenCV学习笔记12 OpenCV图像处理模块ImgProc Module. Sets of parallel lines remain parallel after an affine transformation. The functions in this section perform various geometrical transformations of 2D images. My Algorithm (Bilinear Interpolation) OpenCV RMSE= 2. If you find any difficulty or have any query then do COMMENT below. Following is the input image:. Learn OpenCV with basic implementation of different algorithms. The Affine Transformation is a general rotation, shear, scale, and translation distortion operator. For each, compute the unique affine transformation they define. The Homography transformation is a popular geo-referencing technique used worldwide. How 일 수있는 이유를 찾으려고합니다. Affine transforms can perform any combination of translations, scales/flips, shears, and rotations. The greater the number of matched pairs of points, the greater the accuracy of the estimated transformation. non-uniform scaling in some. The image transformation can be expressed in the form of a matrix multiplication using an affine transformation. OpenCV boasts of an active user base all over the world with its use increasing day by day due to the surge in computer vision applications. Following is the input image:. This is the C function called by the opencv C++ function and as a extra benefit, it return false if the affine matrix cannot be solved. I am an entrepreneur who loves Computer Vision and Machine Learning. So the difference between a $2\times3$ matrix and a $4\times4$ matrix was only from your way of writing it; this works the same way as an affine transform in three dimensions, just with one fewer dimension. An affine transform generates a matrix to transform the image with respect to the entire image. Transform type, specified as one of three character strings. You will see these functions: cv2. gz - computes steerable filters, SIFT descriptor, moments, differential invariants, complex filters, shape context, spin images, PCA-SIFT. Once we have the affine transformation matrix, we use the function to apply this matrix to the input image. Why? Because matrix multiplication is a linear transformation. [Bad (19 instances)] : In these cases Fast-Match did not find the correct area, and the reason isn't one of the above (occluded, out of plane/image). Calculates an affine transform from three pairs of the corresponding points. de <NAME> gabriele. \n" " --ba_refine_mask (mask)\n" " Set refinement mask for bundle adjustment. Here we will learn to apply the following function on an image using OpenCV: Image Transformations - Affine and Non-Affine Transformation; Image Translations - Moving image up, down, left and right; Rotation of image - Spinning the image. M output 2x3 affine transformation [A|b] matrix (see below). Geometric operations performed on an image, changes the spatial relationships of the image pixels. But in some cases, you may need elliptical/circular shaped kernels. That is it will modify an image to perform all four of the given distortions all at the same time. In OpenCV, I can affine transform an image using: M = np. I start to play with OpenCV using Greg Borensteins Processing implementation. These steps include subtracting a mean pixel value and scaling the image. Linux binaries - implementation based on the publications and source code provided by the authors, see publications for details. … and we get a (20,10,10,1) homogeneous vector ! Remember, the 1 means that it is a position, not a direction. Output Input. CS231A Section 6: Problem Set 3 Kevin Wong • 3 matches determine an affine transformation – Guide to building OpenCV Mex files for windows that. The usual way to represent an Affine Transform is by using a 2 \times 3 matrix. Converting from x/y back to row/col uses the inverse of the affine transform. warpAffine and cv2. The first one is used to remove simple geometrical transformations such as rotations, scales, translations, and their combinations, but it can't turn converging lines into parallel ones. The study includes the computation of forces acting on the caissons. objname: The optional name of the node to read (if empty, the first top-level node will be used). Region detectors - Linux binaries for detecting affine covariant regions. An affine transformation is an important class of linear 2-D geometric transformations which maps variables (e. getAffineTransform(). warpAffine(arr, M, (cols, row)) this works if the image is represented as bitmap. I've implemented a PVImage to MAT class to access OpenCV methods direct. The usual way to represent a affine transformation is using a 2 x 3 matrix. Where is it used?. Apply an affine transformation. Sample Code. An affine transformation is composed of rotations, translations, scaling and shearing. gz - computes steerable filters, SIFT descriptor, moments, differential invariants, complex filters, shape context, spin images, PCA-SIFT. opencv based on median filtering method, median filter was proposed in 1971 by Tukey, which is a non-linear filtering techniques and typical. Detectors evaluation - Matlab files to compute the repeatability. Affine transformation - OpenCV 3. Following is the input image:. Affine normalization • Problem: There is no unique transformation from an ellipse to a unit circle • We can rotate or flip a unit circle and it still stays a unit circleWe can rotate or flip a unit circle, and it still stays a unit circle. This post will help you learn how to rotate images using opencv python library. [Bad (19 instances)] : In these cases Fast-Match did not find the correct area, and the reason isn't one of the above (occluded, out of plane/image). OpenCV 3 is a native cross-platform library for computer vision, machine learning, and image processing. to the identity matrix - Then for each simpler transform, T, pre-multiply the matrix. Opencv can be used to rotate the image in any direction that user wants. The following are code examples for showing how to use matplotlib. Demo of Affine transformations with OpenCV. Although it seems little bit complicated, OpenCV provides some built-in functions making it easy to do it. I want to perform the affine transformation on a set of points. 2 Transformations transformations Preservation of affine combinations of points. Intensity Interpolation: Finding the intensity values at the newly calculated positions. An affine transformation has six degrees of freedom including the 2 by 2 affine matrix A and the offset two-component vector t. uses RANSAC framework in calib3d includes accuracy test uses SVD decomposition for solving 3 point equation things to consider: is solving through SVD fast enough? should we make it symmetric and use EIGEN val decomposition? is checking for coplanar points robust enough?. Translation and Euclidean transforms are special cases of the Affine transform. Feature Matching with Affine-Function Transformation Models Abstract: Feature matching is an important problem and has extensive uses in computer vision. I start to play with OpenCV using <NAME> Processing implementation. Learning OpenCV puts you right in the middle of the rapidly expanding field of computer vision. They do not change the image content but deform the pixel grid and map this deformed grid to the destination image. Transform the face for the neural network. Linux binaries - implementation based on the publications and source code provided by the authors, see publications for details. 3) Just do the warping on the CPU side using, say, OpenCV or scipy. edu September 8, 2005 Abstract Geometric transformations are widely used for image registration and the removal of geometric distortion. Given a set of facial landmarks (the input coordinates) our goal is to warp and transform the image to an output coordinate space. cvEstimateAffineTransformation function to wrap the opencv C function defined in the opencv_video module. Detectors evaluation - Matlab files to compute the repeatability. In this video, we'll use OpenCV's built-in functions for warping an image using affine and perspective transformations. In this part, we will cover how to implement how to rotate and shear images as well as bounding boxes using OpenCV's affine transformation features. If any errors are found, please email me at jae. CS231A Section 6: Problem Set 3 <NAME> • 3 matches determine an affine transformation – Guide to building OpenCV Mex files for windows that. In OpenCV an Affine transform is stored in a 2 x 3 sized matrix. 3626099; 0, 0, 1] I would now like to apply the transform to a vector< Pointf >, so that it will transform each point as it would be if they were in the image. opencv based on median filtering method, median filter was proposed in 1971 by Tukey, which is a non-linear filtering techniques and typical. warpAffine and cv2. The size of *this and other must be four. 2D and 3D Transformations <NAME> Adapted from notes by Yong Cao Virginia Tech. warpAffine(arr, M, (cols, row)) this works if the image is represented as bitmap. The default is ray. OpenCV (Open Source Computer Vision Library) is an open source computer vision and machine learning software library. Therefore, I think estimateAffine3D can estimate a affine transformation includes true 3D scaling/shearing. A backwards transformation? Why aren't we mapping from the input image to the output? For each point maintain the new continuous coordinates. Affine Transformation in Image Processing: Explained with C++ Transformations are used to change the geometry of the contents within the image. So my second image is obtained in two steps, Resize the first image to 0. I know some (at least 3) points on the current image and the same points on the "new image" (after the affine transformation). By voting up you can indicate which examples are most useful and appropriate. • A composite transformation matrix is easily constructed from a series of more simple transformations - Initialize. The usual way to represent an Affine Transform is by using a matrix. affine_transform, and then transferring the data to the GPU for the rest of my CUDA pipeline The source image is 1024 x 768 single channel, getting resampled down to 64 x 80 using a transformation matrix that gets computed online. Then the transformations are then applied on the model and the results show that the model's shape is changed (i. You will see these functions: cv2. Use the OpenCV function cv::getRotationMatrix2D to obtain a $$2 \times 3$$ rotation matrix; Theory. OpenCV provides two transformation functions, cv2. The functions in this section perform various geometrical transformations of 2D images. Given a set of facial landmarks (the input coordinates) our goal is to warp and transform the image to an output coordinate space. From the above, We can use an Affine Transformation to express: Rotations (linear transformation) Translations (vector addition) Scale operations (linear transformation) you can see that, in essence, an Affine Transformation represents a relation between two images. In this video, we'll use OpenCV's built-in functions for warping an image using affine and perspective transformations. Lecture 2: Geometric Image Transformations Harvey R<NAME>. pdf - Free ebook download as PDF File (. Only the following transformations are available Only the following transformations are available theta, tx, ty, shear, zx, zy, flip_horizontal, flip_vertical, channel_shift_intensity, brightness. robot Sep 17 '14 at 14:17. A backwards transformation? Why aren't we mapping from the input image to the output? For each point maintain the new continuous coordinates. Assume x 1 and x 2. Pure numpy and matplotlib was used to give a low-level intuitive description of how affine transformations work. I've got a Affine transform matrix in OpenCV from the KeypointBasedMotionEstimator class. "This library is useful for practitioners, and is an excellent tool for those entering the field: it is a set of computer vision algorithms that work as advertised. Although it is written in optimized C/C++, it has interfaces for Python and Java along with C++. The following shows the result of a affine transformation applied to a torus. I am an entrepreneur who loves Computer Vision and Machine Learning. We scale the image by half and rotate it by 90 degrees anticlockwise. In fact, to avoid sampling artifacts, the mapping is done in the reverse order, from destination to the source. Images and OpenCV. Affine transformation - OpenCV 3. Post Organization. Your report should include: (i) the transformed image 1 and (ii) the. My idea is to calculate the AffineTransformation matrix (through the OpenCV function: GetAffineTransform) using 3 matching features found by SURF. In OpenCV, I can affine transform an image using: M = np. Translation and Euclidean transforms are special cases of the Affine transform. OpenCV Perspective Transform. transformation matrix are the images of the base vectors! That gives us an easy method of nding the matrix for a given linear transformation. I know some (at least 3) points on the current image and the same points on the "new image" (after the affine transformation). warpAffine()およびcv2. • T = MAKETFORM('affine',U,X) builds a TFORM struct for a • two-dimensional affine transformation that maps each row of U • to the corresponding row of X U and X are each 3to the corresponding row of X. Hence in affine transformation the parallelism of lines is always preserved (as mentioned by EdChum ). The following are code examples for showing how to use skimage. Affine transformations In this section, we will discuss about the various generalized geometrical transformations of 2D images. For color conversion, we use the function cv2. Where is it used?. [Bad (19 instances)] : In these cases Fast-Match did not find the correct area, and the reason isn't one of the above (occluded, out of plane/image). 何为仿射变换(Affine Transformation) 9. In fact, to avoid sampling artifacts, the mapping is done in the reverse order, from destination to the source. warpAffine(arr, M, (cols, row)) this works if the image is represented as bitmap. Calculates an affine transform from three pairs of the corresponding points. They are extracted from open source Python projects. An affine transformation is a 2-dimension cartesian transformation applied to both vector and raster data, which can rotate, shift, scale (even applying different factors on each axis) and skew geometries. Learning OpenCV puts you right in the middle of the rapidly expanding field of computer vision. In this chapter you will learn about two important classes of geometric image transformations—Affine and Perspective—and how to represent and use them as matrices in your code. The corners • may not be collinear. Affine and Perspective Transformation 02/05/2017 02/05/2017 physiophile openCV , python In affine transformation ( link , link2 ), all parallel lines in the original image will still be parallel in the output image. ) in the field. The Estimate Geometric Transformation block supports Nonreflective similarity, Affine, and Projective transformation types, which are described in this section. An affine transform produces parallelograms (which includes standard rectangles). This is a port of Korman et al implementation of Fast Affine Template Matching in C++. by <NAME> February 15, 2018. 1) transform vertices to the coordinate system of the camera, 2) transform the result to the 2D plane of the screen, 3) draw triangles using the resultant 2D vertices. PythonバインディングされたOpenCV関数をコールするサンプルスクリプトを掲載。 アフィン変換(Affine Transformation). Python, OpenCVで画像の幾何変換（線形変換・アフィン変換・射影変換）を行うには関数cv2. Notes-----Transformations. getAffineTransform(). 仿射变换 affine transformation ; 4. 2D piecewise affine transformation. I have a dozen years of experience (and a Ph. That means that to simulate a camera transformation, you actually have to transform the world with the inverse of that transformation. In this section we will perform simple operations on images using OpenCV like opening images, drawing simple shapes on images and interacting with images through callbacks. For color conversion, we use the function cv2. Demo of Affine transformations with OpenCV. This example shows how to use the Piecewise Affine Transformation. Section 1 – Image Pre-Processing: In this section, we’ll describe the pre-processing steps that are applied to an input image. 4 with python 3 Tutorial 21 by <NAME> March 7, 2018 We're going to learn in this tutorial how to detect the lines of the road in a live video using Opencv with Python. Essentially the image is multiplied by 2x3 matrix to perform the transformation. We will also discuss the basic of image processing and provide the detail explanation related to the OpenCV functions. Although it is written in optimized C/C++, it has interfaces for Python and Java along with C++. Learn OpenCV with basic implementation of different algorithms. "-<NAME>. , the midpoint of a line segment remains the midpoint after transformation). Lowe, University of British Columbia, came up with a new algorithm, Scale Invariant Feature Transform (SIFT) in his paper, Distinctive Image Features from Scale-Invariant Keypoints, which extract keypoints and compute its descriptors. Carlson Center for Imaging Science Rochester Institute of Technology [email protected] Satya Mallick. GIMP Tutorial – using the Perspective Tool by <NAME> on YouTube. So, a pixel value at fractional coordinates needs to be retrieved. 0011812132, 1. Shows how to do a perspective transform using GIMP. In geometry, an affine transformation, affine map or an affinity (from the Latin, affinis, "connected with") is a function between affine spaces which preserves points, straight lines and planes. Sets of parallel lines remain parallel after an affine transformation. However, existing feature matching methods support either a specific or a small set of transformation models. Note that the matrix form of an affine transformation is a 4-by-4 matrix with the fourth row 0, 0, 0 and 1. The functions in this section perform various geometrical transformations of 2D images. warpAffine(arr, M, (cols, row)) this works if the image is represented as bitmap. non-uniform scaling in some. Linux binaries - implementation based on the publications and source code provided by the authors, see publications for details. Step3: Use RANSAC to robustly fit best affine transformation to the set of point matches. Hence to overcome this problem, we proposed a system using Modified Affine Transformation with Viola Jones. Affine transformation for. opencv学习(6)---仿.


text:
It would be a nice addition if the slings had varying additional damage based on the ammunition

for example the Bows have a base damage range which is increased further with use of different arrow types
Stone-tipped Arrow
Copper-tipped Arrow +3
Bronze-tipped Arrow +5
Iron-tipped Arrow +7
Steel-tipped Arrow +10

perhaps something could be implemented for the Sling ammunition along the same lines
one idea would be to use the existing processed stones like

• Stone
• Rough hewn Stone +1
• Hewn stone +3


text:
# Impedance of R and L in series Calculator

## Calculates the impedance of the resistor and inductor in series.

 Resistor R GΩMΩkΩΩmΩμΩ Inductor L HmHμHnH Frequency f GHzMHzkHzHz 6digit10digit14digit18digit22digit26digit30digit34digit38digit42digit46digit50digit Impedance |Z| Ω Phase φ degree
 $\normal Impedance\ of\ R\ and\ L\ in\ series\\\vspace{5}\hspace{30}Z=R+j\omega L\ ,\hspace{20}\omega=2\pi f\\\hspace{30}|Z|=sqrt{R^2+(\omega L)^2}\\\vspace{10}\\Phase\ difference\\\hspace{30} \phi =\tan^{\tiny -1}{({\large\frac{\omega L}{R}})}\\$

Impedance of R and L in series
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[1]  2019/03/25 19:45   Male / 20 years old level / High-school/ University/ Grad student / - /
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NDT testing
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Product selection to eliminate noise in a control loop
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Personal education
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Induction motor stator resistance effect on low frequency drive
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Inductor''s Inductance measurement.
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Inductor''s Inductance calculation.
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assignment
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Self study
[10]  2013/01/13 23:15   - / 30 years old level / An office worker / A public employee / Very /
Purpose of use
Checking some calculations.
Comment/Request
How to calculate for a circuit as such...
R1 and L in series, and this series is then in parallel with another R2?
Appreciate if you can give me some tips.
Thanks...

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summaryrefslogtreecommitdiffstats log msg author committer range
blob: f30a42082c429fe9424cde41a1ef7d1016caa321 (plain)
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Install isohybrid

\n" msgstr "" #. type: Plain text msgid "" "In **Debian Wheezy**, **Ubuntu 14.04**, and earlier the `isohybrid` utility " "is included in the `syslinux` package. To install it, execute the following " "command:" msgstr "" #. type: Plain text #, no-wrap msgid " sudo apt-get install syslinux\n" msgstr "" #. type: Plain text msgid "" "In **Debian Jessie**, **Ubuntu 14.10**, and later the `isohybrid` utility is " "included in the `syslinux-utils` package. To install it, execute the " "following command:" msgstr "" #. type: Plain text #, no-wrap msgid " sudo apt-get install syslinux-utils\n" msgstr "" #. type: Plain text #, no-wrap msgid "\n" msgstr "" #. type: Plain text #, no-wrap msgid "

Find out the device name of the device

\n" msgstr "" #. type: Plain text msgid "The device name should be something like `/dev/sdb`, `/dev/sdc`, etc." msgstr "" #. type: Plain text msgid "" "If you are not sure about the exact device name, with GNOME, do the " "following:" msgstr "" #. type: Bullet: ' 1. ' msgid "" "Make sure that the USB stick or SD card onto which you want to install Tails " "is unplugged." msgstr "" #. type: Plain text #, no-wrap msgid "" " 1. Open GNOME Disk Utility from the menu\n" " \n" " Applications ▸\n" " Accessories ▸\n" " Disk Utility\n" " \n" msgstr "" #. type: Bullet: ' 1. ' msgid "" "Disk Utility lists all the current " "storage devices in the left pane of the window." msgstr "" #. type: Bullet: ' 1. ' msgid "Plug the USB stick or SD card onto which you want to install Tails." msgstr "" #. type: Plain text #, no-wrap msgid " A new device appears in the list of storage devices. Click on it.\n" msgstr "" #. type: Bullet: ' 1. ' msgid "" "In the right pane of the window, verify that the device corresponds to your " "device, its brand, its size, etc." msgstr "" #. type: Plain text #, no-wrap msgid " [[!img destination_device_description.png link=no alt=\"Drive description containing: 'Device: /dev/sdc'\"]]\n" msgstr "" #. type: Plain text #, no-wrap msgid "" " On this screenshot, the USB stick is a Kingston DataTraveler of 2.0\n" " GB and its device name is `/dev/sdc`. Yours are probably different.\n" msgstr "" #. type: Plain text #, no-wrap msgid "
\n" msgstr "" #. type: Plain text #, no-wrap msgid "" "If you are not sure about the device name, you should stop\n" "proceeding or you risk overwriting any hard disk on the\n" "system.\n" msgstr "" #. type: Plain text #, no-wrap msgid "
\n" msgstr "" #. type: Plain text #, no-wrap msgid "\n" msgstr "" #. type: Plain text #, no-wrap msgid "

Do the copy

\n" msgstr "" #. type: Plain text #, no-wrap msgid "

All the data on the installed device will be lost.

\n" msgstr "" #. type: Plain text msgid "" "Execute the following commands, replacing `[tails.iso]` with the path to the " "ISO image that you want to copy and `[device]` with the device name found in " "step 2." msgstr "" #. type: Plain text #, no-wrap msgid "
\n" msgstr "" #. type: Plain text #, no-wrap msgid "" "

Note that the isohybrid command modifies the ISO image. As a\n" "consequence, you won't be able to [[verify|download/#verify]] it again\n" "afterwards. We recommend you to execute those commands on a copy of the\n" "original ISO image that you downloaded.

\n" msgstr "" #. type: Plain text #, no-wrap msgid "" " isohybrid [tails-isohybrid.iso] -h 255 -s 63\n" " dd if=[tails-isohybrid.iso] of=[device] bs=16M\n" msgstr "" #. type: Plain text msgid "" "Here is an example of the commands to execute, yours are probably different:" msgstr "" #. type: Plain text #, no-wrap msgid "" " isohybrid '/home/amnesia/Desktop/tails-0.6.2-isohybrid.iso' -h 255 -s 63\n" " dd if='/home/amnesia/Desktop/tails-0.6.2-isohybrid.iso' of=/dev/sdc bs=16M\n" msgstr "" #. type: Plain text #, no-wrap msgid "
\n" msgstr "" #. type: Plain text #, no-wrap msgid "" "If you are not sure about the path to the ISO image or if you get a\n" "No such\n" "file or directory error message, you can first type\n" "dd, followed by a space, and\n" "then drag and drop the icon of the ISO image from a file browser onto the\n" "terminal. This should insert the correct path to the ISO image in the terminal.\n" "Then complete the command and execute it.\n" msgstr "" #. type: Plain text msgid "" "If you don't see any error message, Tails is being copied onto the device. " "The whole process might take some time, generally a few minutes." msgstr "" #. type: Plain text #, no-wrap msgid "
\n" msgstr "" #. type: Plain text #, no-wrap msgid "" "

Once the command prompt reappears, you can shutdown your computer, and\n" "[[start Tails|start_tails]] from this new device.

\n" msgstr "" #. type: Title = #, no-wrap msgid "Troubleshooting\n" msgstr "" #. type: Title ### #, no-wrap msgid "dd: /dev/sdx: No such file or directory" msgstr "" #. type: Plain text msgid "" "Then double-check the name of the device you found in [[step 2|" "linux#step_2]]." msgstr "" #. type: Title ### #, no-wrap msgid "dd: /dev/sdx: Permission denied" msgstr "" #. type: Plain text msgid "" "You might also have committed a mistake in the device name, so please double-" "check it. If you are sure about the device name, this could be a permission " "problem and you could need to gain administration privileges before running " "the commands in the terminal. That could be:" msgstr "" #. type: Plain text #, no-wrap msgid "" " isohybrid [tails.iso] -h 255 -s 63\n" " sudo dd if=[tails.iso] of=[device] bs=16M\n" msgstr "" #. type: Title ### #, no-wrap msgid "dd: tails.iso: No such file or directory" msgstr "" #. type: Plain text msgid "" "Then you surely have committed a mistake on the path to the ISO image in " "[[step 3|linux#step_3]]." msgstr ""