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Easy math/physics questions about kinematics

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Hi, Okay, I''m working on a program that will output RIB code for a ball that, at time zero, is sitting on a table. Someone hits the ball (force is parallel to table surface) and it travles off table & bounces on ground. Anyway, I have a few general quick & easy questions about some stuff: My program will have a "start" point, "end" (where it''s going to next) and a "force" variable. I am assuming that I can do the following: vector velocity = force * (end-start); //where end, start & velocity are vectors) First, will vector velocity be in units/sec after I do the previous calculation? At time 1, how do I calculate the balls new position? (related: what is a displacment function for this?) I understand most concepts covered in calculus, and some _very_ basic physics concepts. Thank you, -Pete

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I think I might have figured out the displacement functino I was in search of. . . can anyone back me up on this:

Displacement = d

d(2) = d(1) + v(1)*t + ((a*t^2)/2)

Does that seem right?

-pete

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Yes, for constant acceleration. It also handles deceleration (just use a negative value for accelration), and works in 2 and 3D if you replace d, v and a with vectors. The main thing is the acceleration must be costant for such a formula to work.

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the second post had a known and usuable equation for constant acceleration.

the first post had this:
I am assuming that I can do the following:
vector velocity = force * (end-start); //where end, start & velocity are vectors)



always replace numbers with their units and ask yourself does this "equal" in the sense that the units are the same.

you got a velocity = force * velocity. doesnt equal. its an invalid equation. physics will always mandate valid equations have units that equal.



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in regards to that second equation. that has acceleration which is a.

a = f/m

accleration equals force divided by mass. if your going to have a concept of force in your simulation youll need a mass. BUT!! thats force applied constantly over time. your force from the side isnt being applied constantly over time.

easier to start with is just dropping a ball. then you have a force constantly over time with a rest start state. to use the idea of a "bump" you got a concept of an "impulse" its different. in practice though for you easier then that is a sort of cheap v = a*t over a small t. all this says is that im applying a force over a small time. im pushing for 1/10 of a second. then from there you have the easy x equation.

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