Kinematics and rigid body simulation

[clipmon]

Hello, I have a question regarding how I'm simulating motion using the common time integrating approach to simulating rigid bodies.

I have net forces and solve for position from those:
acceleration = force / mass
velocity = initial velocity + acceleration * time
position = initial position + velocity * time

but if you combine the above and distribute you get something that matches the kinematic equation, but is missing the one half term. Where'd it go?

position = initial position + (initial velocity + (force / mass) * time) * time)
position = initial position + initial velocity * time + (force / mass) * time^2

which does not equal the kinematic equation
y = y0 + v0*t + (1/2)a*t^2

Where'd the one half go?

 

[graphene]

position = initial position + velocity * time

That applies only if the velocity is constant. Since you have an acceleration, velocity is not constant & you cannot use these expressions.

Do you know calculus? (simple integration). It can very easily be proved with calc.

 

[Cleonis]

As graphene points out, the velocity isn't constant.

Here is a short route to arriving at y = 1/2*a*t^2

If the velocity is constant then in diagram the distance covered is proportional to the area of a rectangle. Draw a diagram with time on the horizontal axis and velocity on the vertical axis. So the line representing the velocity at each point in time is horizontal. As time progresses the rectangle stretches in horizontal direction. In other words, the distance covered is given by the area that is enclosed between the horizontal axis and the line that represents the velocity.

Now the case of uniform acceleration.
Then the line representing the velocity has a uniform slope. Interestingly, the total distance covered is still the area that is enclosed between the horizontal axis and the line that represents the velocity. The enclosed area is a rectangular triangle, and the surface area of a triangle is (1/2)*width*height.

At t=0 the velocity is 0,
At t_end the velocity has climbed to a*t_end

The area of the enclosed triangle is (1/2)*a*t_end^2

 

[clipmon]

These equations should be good for constant acceleration, and since I'm doing fixed time steps changes acceleration should always be constant. I've taken calculus and physics and have the books and follow the derivation of the kinematic equation.

y'' = a
y' = a*t + c1
y = (1/2)a*t^2 + c1 * t + c2

where c1 = initial velocity and c2 = initial position

Looking in my physics book I see they define the kinematic equation in terms of average velocity

avgvel = (initial velocity + velocity) / 2

Which I think is where the one half is coming from, but do I need to find average velocity if I'm doing 16ms time steps? I'll start looking into the difference of average velocity / velocity to see if I can figure things out.

 

[clipmon]

Alright Cleonis's explanation relating it to triangle area made things mentally click for me, I need to use the average velocity term. Thanks!

 

[Cleonis]

Looking in my physics book I see they define the kinematic equation in terms of average velocity
avgvel = (initial velocity + velocity) / 2

By coincidence a calculation that works with average velocity can end up looking the same. Don't be distracted by that: the relation:

y = (1/2)*a*t^2

is exact. Whenever the acceleration is uniform it is exact for every point in time.

Your purpose is not to average the velocity; your purpose is to use the exact velocity at each point in time. In this particular case it ends up looking the same.

 

[clipmon]

I wrote this paper and example code this weekend. I'm more of a programmer than a writer / physicist, so if you could look over those and correct my math, grammar, and give me any input I'd appreciate it. The code works, but I'd like to improve it and my understanding of the physics behind it. To compile the example code you may need to download glew. http://glew.sourceforge.net/

http://students.csci.unt.edu/~akw0088/rigid.pdf
http://students.csci.unt.edu/~akw0088/rigid.zip

Thanks!