does a re-trigger on a basic rule reset a wait timer?
trying to migrate a very simple rm to basic rule.

rm 4 rule:

[[image]](https://community.hubitat.com/uploads/default/original/3x/f/6/f6ca33d1b381de4ca2ea35595e4f5a30dd94824d.png "image")

basic rule:

[image]

can't see time limits i set, but they're same. shouldn't this do the same? only part i'm not sure about is rm rule will reset the "wait" if motion continues... i'm not sure if this one does.

thanks!
i think here: [[released] basic rules](https://community.hubitat.com/t/released-basic-rules/68942)

my question is answered by:

“ both wait for minutes and wait until are cancelled if the rule is triggered again by it's primary trigger.”
You must enable only one timer at a time.  

Timer1 to go up.  

Timer2 to go down.



When any Timer reaches the destination floor, it shoud be self disabled.



On both timers you check for floor 4. IMO you should chek if the current floor is the destination or greater on timer1 and the destination or lower in timer2.



Hope this helps [image] ryan\_blything:

i have difficulty understanding why (since ultimately, rts should be dependent on keyboard press rather than frame rate).

builder experiments are structured around a drawing loop and event cycle that is driven by the screen refresh rate. so the keyboard is checked once … A trigger is not what you want.


What you need is a scheduled job.


Take a look at the official documentation on the subject:


[Scheduling Jobs with Oracle Scheduler](http://docs.oracle.com/cd/B28359_01/server.111/b28310/scheduse.htm#i1033533)


Feel free to ask more questions if you find any trouble when creating your scheduled job. I don't know that printer. But Arduino has a feature that a USB connection causes a reset. The Idea is that this helps when doing software update as the Arduino Boot loader will be active for one second after that reset.

 This can be disabled, but needs hardware modification. never mind, finally solved it. i called this.engine.runrenderloop while it’s already running, therefore causing the speed to run twice, because it added a loop/callback the 2nd time.

all i need to do for the ‘resume’ part of my function is to always call this.engine.stoprenderloop() before i call t… hi [@interactiverendering](/u/interactiverendering) ,

this is a little more complicated than it appears to be. in order to remove the close button from the form (because it would make revit crash in some cases) i had to remove that logo too. i’ll try and bring it back and let you know :slight\_smile: hi there,

i am making some assumptions about the requirements of your experiment here but lets see if this helps.

short break

would you like the short break to be at the start of each trial ? i.e. the inter-trial-interval? if so try the following:

add a code component to your “trial” routine (c… its to do with the jit optimisation, the same is true if you insert conditional debug statement in the code, matlab needs to prepare to stop execution. hi there!we are asked to determine the reaction order of a and b using the given data.recall that the rate law only focuses on the reactant concentrations and has a general form of:rate= kaxbyk = rate constanta & b = reactantsx & y = reactant ordersreactant ain order to calculate the order of the reaction with respect to a, we must look for the experimental data where the concentration of a changes while the concentration of b remains the same.trial[a] (mol/l)[b] (mol/l)initial rate (m/s)10.1000.100520.3000.1004530.1000.2001040.3000.20090the rate laws are:rate= kaxbytrial 1:5 m/s= k0.100x0.100ytrial 2:45 m/s= k0.300x0.100ysolve for x using the ratio of the two rates:***the larger rate should be on the numeratorrate 2rate 1=45 m/s5 m/s= k0.300x0.100yk0.100x0.100y• for reactions at a constant temperature, the rate constant remains the same. thus, it could be canceled out.• since the reaction order ‘y’ of b is also the same, it could also be canceled out.45 m/s5 m/s= k0.300x0.100yk0.100x0.100y9 = 3xlog 9=x·log3log 9log3=x·log3log3x=2the reaction order, x, with respect to a is 2. reactant bin order to calculate the order of the reaction with respect to a, we must look for the experimental data where the concentration of b changes while the concentration of a remains the same.trial[a] (mol/l)[b] (mol/l)initial rate (m/s)10.1000.100520.3000.1004530.1000.2001040.3000.20090the rate laws are:rate= kaxbytrial 1:5 m/s= k0.100x0.100ytrial 3:10 m/s= k0.100x0.200ysolve for x using the ratio of the two rates:***the larger rate should be on the numeratorrate 3rate 1=10 m/s5 m/s= k0.100x0.200yk0.100x0.100y• for reactions at a constant temperature, the rate constant remains the same. thus, it could be canceled out.• since the reaction order ‘x’ of a is also the same, it could also be canceled out.10 m/s5 m/s= k0.100x0.200yk0.100x0.100y2 = 2ylog 2=y·log2log 2log2=y·log2log2y=1the reaction order, y, with respect to b is 1. [image] dueck456:

is there an easy way to detect when new code is flashed to the device so i can reinitialize the retained variables?

below is a link to a post by [@rickkas7](/u/rickkas7) about storing retained data in a struct.

[image] dueck456:

saving a know value to eeprom jumps to mind

there is …