Kinematics and rigid body simulation

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Discussion Overview

The discussion revolves around the simulation of motion in rigid body dynamics, specifically focusing on the integration of forces to determine position, velocity, and acceleration. Participants explore the relationship between these variables and the derivation of kinematic equations, addressing the apparent absence of the one-half term in their calculations.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant questions the absence of the one-half term in their derived position equation compared to the kinematic equation.
  • Another participant clarifies that the equation for position derived from constant velocity is not applicable when acceleration is present, suggesting the use of calculus for proper integration.
  • A participant explains the relationship between area under the velocity-time graph and distance traveled, emphasizing that for uniform acceleration, the area forms a triangle, leading to the one-half factor.
  • One participant asserts that their equations are valid for constant acceleration and references their understanding of kinematic equations, noting the role of average velocity in deriving the one-half term.
  • A later reply acknowledges the clarity brought by the triangle area analogy and confirms the need to consider average velocity in their calculations.
  • Another participant emphasizes that the relation y = (1/2)*a*t^2 is exact for uniform acceleration and cautions against confusing average velocity with instantaneous velocity.
  • A participant shares a paper and example code related to their simulation work, seeking feedback on both the mathematical and programming aspects.

Areas of Agreement / Disagreement

Participants express differing views on the necessity of using average velocity versus instantaneous velocity in their simulations. While some agree on the importance of the one-half term in the context of uniform acceleration, others highlight the distinction between average and exact velocities, indicating that the discussion remains unresolved regarding the best approach for simulation.

Contextual Notes

Some participants reference calculus and the derivation of kinematic equations, but there are unresolved aspects regarding the application of average versus instantaneous velocity in simulations, particularly with fixed time steps.

Who May Find This Useful

This discussion may be useful for programmers and physicists interested in rigid body dynamics, simulation techniques, and the mathematical foundations of kinematics.

clipmon
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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?
 
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clipmon said:
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.
 
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 tend the velocity has climbed to a*tend

The area of the enclosed triangle is (1/2)*a*tend^2
 
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.
 
Alright Cleonis's explanation relating it to triangle area made things mentally click for me, I need to use the average velocity term. Thanks!
 
clipmon said:
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.
 
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