Numerically determine area and circumference

In summary, the conversation is about determining the area and circumference of a ball that is being thrown in a small game. The problem is presented and the equation for a cardioid is mentioned as a possible solution. The path of the ball is described as a distorted ellipse/circle and someone suggests asking for help from astrophysicists or cosmologists. The conversation then goes on to discuss integrating in polar coordinates and obtaining a solution for the area of the ball.
  • #1
anders_svensson
1
0
Hey guys

I'm currently making a small game (simple space game) and have a problem with numerically determine area and circumference of a ball that is being thrown.

The problem is presented here:

http://trasigkondensator.tripod.com/ball.htm

Thanks in advance
 
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  • #2
That looks like a cardioid. It's general equation in polar coords is: r=a(1-cost), but it could also be r=a(1+cost), i.e. flipped around. I think the maximum distance from the point where it has an 'indent' is 2a, but I'm not entirely sure. a could possibly be either the length of the rope or the radius of the circle.

The path of motion of the ball looks like a distored ellipse/circle. Perhaps someone who's into cosmology can help you out.

Sorry I couldn't be of much help.
 
  • #3
I think you meant topology, not cosmology. :)
 
  • #4
Topology? I don't know much about topology, but I thought some of the astrophysicists/cosmologists could help because the path of the ball looks like several elliptic and circular orbits. :shy:
 
  • #5
Oh, I see.
 
  • #6
Assuming devious is right. Then it would just be a matter of integrating in polar coordinates. Just get a function r in terms of [itex]\theta[/itex] that describes the distance of the ball from the center of the circle and integrate:
[tex]A=\int_0^{2\pi}\int_0^{r(\theta)}r'dr'd\theta[/tex]
[tex]A = \frac{1}{2}\int_0^{2\pi}r^2(\theta)d\theta[/tex]
edit - looked up a cardioid, a = the radius of the circle. So:
[tex]A = \frac{1}{2}\int_0^{2\pi}a^2(1-cos(\theta))^2d\theta[/tex]
[tex] = \frac{1}{2}a^2\int_0^{2\pi}sin(\theta)^2d\theta[/tex]
[tex] = \frac{1}{4}a^2\int_0^{2\pi}(1-cos(2\theta))d\theta[/tex]
[tex] = 3a^2\pi[/tex]
 
Last edited:

1. How do you numerically determine the area of a circle?

Numerically determining the area of a circle involves using the formula A = πr2, where A is the area and r is the radius. This formula can be applied to any circle, regardless of its size or units of measurement.

2. Can you explain the concept of circumference and how to calculate it numerically?

Circumference is the distance around the edge of a circle. To numerically determine the circumference of a circle, you can use the formula C = 2πr, where C is the circumference and r is the radius. This formula is derived from the relationship between the circumference and the diameter of a circle, which is C = πd.

3. Are there any other methods for numerically determining the area and circumference of a circle?

Yes, there are other methods such as using numerical integration or using a ruler to measure the diameter and then calculating the area and circumference using the formulas mentioned above. However, these methods may not be as accurate as using the formula A = πr2 and C = 2πr.

4. How do you handle decimal values when calculating the area and circumference of a circle?

When calculating the area and circumference of a circle, it is important to use the exact value of π, which is approximately 3.14159. This will ensure that your calculations are as accurate as possible. If you are using a calculator, make sure to use enough decimal places to maintain accuracy.

5. Can you provide an example of numerically determining the area and circumference of a circle?

Sure, let's say we have a circle with a radius of 5 inches. To determine the area, we would use the formula A = π(5)2 = 25π square inches. To determine the circumference, we would use the formula C = 2π(5) = 10π inches. So the area is approximately 78.54 square inches and the circumference is approximately 31.42 inches.

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