2 arc-related problems: finding volume of game controller

matt000r000

Junior Member
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Mar 18, 2009
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i was playing with my old joystick videogame when i was struck with a simple question: what was the volume of the joystick? so i took the problem apart.

the volume is equal to the volume of the spere plus the volume of the cylender minus the sliver of the sphere overlapped by the cylender. no-biggie, i thought.

boy was i wrong. the little sliver of a spere proved dificult. at this point, the actual measurements i took don't matter as much as the way to find the answer. so i figured i could probably simplify the problem to finding the area of a chord. i spent a week pondering this (doing my mathmatical work on paper-plates since i was on vacation in FL, so there was no appropriate sized paper. thats why i can't show all my work) and came up with something, but it accidentaly got covered with ketchup.

anyway, i didn't need to redo my work for it was based on finding the area of the pie-slice formed by the chord, and then subtracting the tirangle between the cord and the center. i realized this would not work in 3-dimentions because a circle can be split into pies, but a sphere can't! so much for a week of vacation. but, as soon as i relized this, i could not figure any other way to approach the problem.

so now i am turning to this site. any help whatsoever is appreciated greatly, on one conditon: i would like to see the work involved if possible. thanks!!

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"if you can't fly, run. if you can't run, walk. if you can't walk, crawl, but by all means keep moving!"
-marthin luther king, jr.
 
Re: 2 arc-related problems...

btw, i forgot, but the 2nd arc problem was: how do you find the area of a chord, knowing the length of the chord and the hight of the arc?
 
Re: 2 arc-related problems...

btw, i forgot, but the 2nd arc problem was: how do you find the area of a chord, knowing the length of the chord and the hight of the arc?


There are three formulae we can use.

The length of the chord is known: \(\displaystyle LC=2Rsin(\frac{\theta}{2})\)

The middle ordinate or 'height of the arc' as you called it: \(\displaystyle MO=R(1-cos(\frac{\theta}{2}))\)

Plug in your knowns for LC and MO, then solve for theta and R.

Next, sub those into the formula for the area of a circular segment(what you called the 'area of the chord'):

\(\displaystyle A=\frac{1}{2}R^{2}({\theta}-sin({\theta}))\)
 
thanx. but do you have the volume of the sphere-slice(i don't know what else to call it)?
 
I am totally sure what you mean. Perhaps a spherical cap or a spherical wedge.

The cap is just that. It's a section of the sphere if we sliced it with a plane at some point.

A spherical wedge is like a wedge cut out of a sphere instead of a circle(like a pie).

Which one?.
 
matt000r000 said:
thanx. but do you have the volume of the sphere-slice(i don't know what else to call it)?

For the volume of a portion of a sphere from \(\displaystyle h=h_o\) to \(\displaystyle h=h_1\):

\(\displaystyle \int_{h_0}^{h_1} \pi r^2 dh\)

The r will vary with h (R=radius of sphere, r=radius of changing circle):

\(\displaystyle h^2 + (R-h)^2 = r^2\)

Implies:

\(\displaystyle 2h^2 -2Rh + R^2 = r^2\)

So,

\(\displaystyle \int_{h_0}^{h_1} \pi r^2 dh = \int_{h_0}^{h_1} \pi (2h^2-2Rh+R^2) dh\)

This integrates to:

\(\displaystyle \pi(\frac{2}{3}h^3 - Rh^2 + R^2h) |_{h_0}^{h_1}\)

To test, try from -R to R (we should get the volume of a sphere)

\(\displaystyle [\pi(\frac{2}{3}R^3 - R^3 + R^3) ] - [\pi(-\frac{2}{3}R^3 + R^3 - R^3) ] = \frac{4}{3} \pi R^3\)
 
thanks.......i guess...... i hav no clue how an integral works..... im prettey shure it has to do with an infinant sequence something like:
4N+6N
----------
3N
as N approaches infinity, but thats all ive been able to gleam of calculus from algebra 1. i also know it is somehow used in finding the area of curves, using an extreme method of exaustion... can some one clear up what an integral is, and what the syntax is(how to use/write it)? or do i have to go to the calculus forum?
 
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