Solved

Elliptic Arc Path

Posted on 2003-11-18
5
344 Views
Last Modified: 2008-03-10
Hi all,

I'm trying to code the "bouncing movement" of a ball for a program.

What I need is the formula from which I can get an exact point (position of the ball) in an elliptic arc. (I couldn't yet find it)

Could anybody please help?

Thanks in advanced,
Techfreelance
0
Comment
Question by:techfreelance
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
5 Comments
 
LVL 22

Accepted Solution

by:
grg99 earned 100 total points
ID: 9771211
If you're trying to simulate a ball falling and then bouncing off the floor, you can try this technique:

Give the ball a starting velocity in the X direction, say 1 m/sec

Give it a starting velocity in the Y direction, if we assume it's being thrown upward, say 0.5 m/sec

Now let's assume there's no air resistance, so the ball is not going to change speed in the X direction,
i.e. X Accelleration equals zero.

But there's gravity, which is a constant downward acceleration, so each second it's going to increase in Y velocity by G m/sec,
on Earth's surface G is IIRC 9.8 meters per second.

So that shoul dbe all you need to calculate the ball's trajectory.  Starting at X,Y position of say (0,0), calculate its position 0.1 second later, easily done since you know the X and Y velocities.  Also calculate a new X and Y velocity, given the X and Y accelerations.  Very simple math.


And Oh, when the ball hits the ground (Y = 0), it will bounce, assume a very elastic ball, so the Y velocity reflects.  Very easy, very simple, as the old Chef Tell used to say,

0
 
LVL 1

Assisted Solution

by:Big_B
Big_B earned 50 total points
ID: 9771284
Not a formula for an elliptical arc but perhaps you try something similar to this.
http://www.brainycreatures.co.uk/physics/friction.asp
0
 
LVL 31

Assisted Solution

by:GwynforWeb
GwynforWeb earned 100 total points
ID: 9771506
If a ball takes off at an angle theta and speed V then the path it takes is

                       x(t) = V*t*cos(theta)

                       y(t) = V*t*sin(theta) - g*t² /2

The time for the ball to come down is  (2V/g)sin(theta) so the above formula is for  0 < t < (2V/g)sin(theta).

I am not sure exactly how much detail you want but there is a start, let me know what else you need.
0
 
LVL 1

Expert Comment

by:Urhixidur
ID: 9838127
To the previous comments I'll simply add that what you're looking at is *not* an elliptical arc, but a parabolic arc.
0

Featured Post

Free Tool: Path Explorer

An intuitive utility to help find the CSS path to UI elements on a webpage. These paths are used frequently in a variety of front-end development and QA automation tasks.

One of a set of tools we're offering as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

A Guide to the PMT, FV, IPMT and PPMT Functions In MS Excel we have the PMT, FV, IPMT and PPMT functions, which do a fantastic job for interest rate calculations.  But what if you don't have Excel ? This article is for programmers looking to re…
Introduction On a scale of 1 to 10, how would you rate our Product? Many of us have answered that question time and time again. But only a few of us have had the pleasure of receiving a stack of the filled out surveys and being asked to do somethi…
This is a video describing the growing solar energy use in Utah. This is a topic that greatly interests me and so I decided to produce a video about it.
Although Jacob Bernoulli (1654-1705) has been credited as the creator of "Binomial Distribution Table", Gottfried Leibniz (1646-1716) did his dissertation on the subject in 1666; Leibniz you may recall is the co-inventor of "Calculus" and beat Isaac…

628 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question