# Another ribbon problem!

Suppose,you have a ribbon that can be tightly wrapped around the equatorial circumference of the earth. Now you increase its length by 1 metre. Now your job is to calculate how much high equally it can be raised all around the circumference of the earth.
I WANT THE EXACT ANSWER.BEST OF LUCK!!! CLUE: The answer is a surprising one.

Note by Bhargav Das
5 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

$$\Delta L=2 \pi \Delta r$$, where $$\Delta r$$ is the desired uniform increase in height and $$\Delta L$$ is the increase in ribbon length. This is the same equation to be used in the other ribbon problem.

I used to ponder this question while running around a racetrack, intrigued how the extra distance between tracks is independent of the radius of curvature. A better question is to ask, at what latitude relative to the equator will the height increase by 1 meter for a stretched ribbon at constant length?

- 5 years ago

what ans. u got?

- 5 years ago

Give me any planet or sphere in the universe and $$\Delta L = 1$$ m, and the height will be $$\Delta r = \frac {1} {2 \pi}$$ m.

- 5 years ago

- 5 years ago

1/2 pie

- 5 years ago

@Leo,correct.

- 5 years ago

1/2 pie

- 3 years, 10 months ago
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 1, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.9893831610679626, "perplexity": 4329.830772576819}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.3, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2018-17/segments/1524125945584.75/warc/CC-MAIN-20180422100104-20180422120104-00366.warc.gz"}