A rope 1 cm wide and 100 m long is wound up. What is the diameter

420

>>8092838
sqrt(10000/pi) cm

>>8092870
are you sure

>>8092838
10000 centimeters squared

>>8092896
Derp 10,000 is the area, so divide by 2, get sqrt(10,000/pi) centimeters

>>8092838
How tightly and in what winding pattern is it wound anon?

>>8092838
Instead of actually figuring this out, I'm gonna try to get a rough estimate because I'm bored. Maybe someone smarter than me can finish the job. So instead of using a spiral lets use concentric circles, with the center circle being 1 cm in diameter. So this first circle takes off ~ 1 cm off the total, so we are left with 99.99 m. The next circle will have a diameter of 3 cm, and a circumference of 3pi cm. This leaves a total length of 99.98575 m.

If we continue this pattern, we can bring the total length down to 0. So for the nth concentric circle, the diameter will be (2n-1) cm. The circumference of the nth circle is then (2n-1)*pi cm. And so we just have to find the k that satisfies: [eqn] 0=9999- \sum_{n=2} ^{k} (2n-1)*pi [/eqn]
And then plug this k value into the equation for the diameter (2k-1) cm. I went ahead and did this on wolfram alpha and got [math] \sqrt{ \frac{9999+pi}{pi} } [/math] or ~56 cm. So it turns out you can pack 100 meters of rope into a spiral of about a half meter diameter.

>>8092917
Loosely, such that a cylinder containing the wound rope has precisely 50% rope and 50% air by mass. In an even equidistant spiral, somewhat similar to OP pic related.

>>8092838
I guess I understand the answer now

The area of the circle has to be the area of the rope, so the area is 10,000cm^2...

I thought I had come up with a cooler question, but that is just boring

Asking for diameter instead of radius is truly a cruel thing to do.

>>8092935
Yea woops
>sqrt(10000/pi) cm
is the radius

>>8092922
The rope is 1x10000 centimeters you dumbass, that won't change if you wrap it into a circle

>>8093048
No need to call names you piece of shit. I was doing an approximation because spirals cannot have a constant diameter. So instead of reposting what everyone else said in the thread, why don't you figure out the diameter as a function of the distance to the end of the rope for me. Oh wait, the only math you can do is how to find areas of rectangles and circles, you sure are a dumbass. Fuck you.

>>8093063
>being buttblasted because you can't see obvious solutions to problems

Kek stay mad

>>8093048
ok smartie solve this one
>>8092928

>>8093246
couldn't find the diameter as a function of distance to the end of the rope I see. Well I hope you at least tried and failed instead of just being a shitposting faggot.

>>8092928
50% rope and 50% air by mass? Don't you mean by volume?

If by volume, then it would be 2 times the diameter wouldn't it?

>>8092922
Damn, you were very close to what I did, basically the same thing, but with integrals instead of summations.

[eqn]10000 = 2\pi \int_1^x r dr[/eqn]
[eqn]10000 = 2\pi \left(\frac{x^2}{2} - \frac{1}{2}\right)[/eqn]

Solve for x in WolframAlpha, and I got [math]\sqrt{\frac{10000 + \pi}{\pi}}[/math]

>>8093366
btw, x equals the radius, so the diameter would be double of that

>>8093366
Should be from 0 to x, and then you get the normal area of a circle, pi*r^2