Physics/ Math people

I was trying to figure out the optimal Force output of our kicker by doing some math. If you increase the weight at the end, then the torque needed is higher, speed lower, and vice versa. So after trying to do the math, I had some questions ( i have taken calcI, but not calc based physics yet).

If you have a position formula and take the first derivative, you get linear velocity, the second derivative is linear acceleration. How would you compute this for a rotational device such as a pendulum or a kicker?

And how would you come up with one position formula for a device like that if x = cosQ * r and y = -sinQ * r ( r being the arm length)(Q = degrees in radians)?

Thanks.

Could you do it with potential energy? Find the potential energy of the spring at the beginning and end of it’s travel (you probably won’t expend it all - I know our kicker stops short of the full spring extension) and the ∆PE is (almost) all transformed into KE.

∆PE is equal to the work done to draw it back (Fdistance), which is also the integral of the graph force over distance (F=-kx, for a perfect spring). If you know how much force it takes to pull your kicker back the distance to the point you want, then that (Fd) would be equal to (1/2)m(v final)^2.

To those who are more experienced with physics than I: was I making this up out of the whole cloth? It seems reasonable to me, but I’m guessing.

This isn’t so much a physics question as it is a testing question. We built a kicker that could vary its height, point of connection on the ball, force applied, and weight of the kicker. Basically we found for our kicker that a length of about 16 inches was the optimal length and more Latex was roughly directly proportional to kick distance and weight had little effect on the distance of the kick.

Please read last paragraph first.

First of all, I’m going to assume your just working with radial motion.(your kicker only moves about an axis without shifting along it)

Now, the equivalent of position is angle. Instead of measuring in feet or such, you will be measuring this in rotations or radians. This makes the equivalent of linear velocity to be angular velocity(often symbolized in formulas as omegaω). It can be measured in rpm or radians per second. The equivalent of linear acceleration is angular acceleration(often symbolized in formulas as alphaα).

Ok, your sort of right on your idea about x and y. However, your trying to work in Cartesian coordinate system(and if you are going to you would have to consider this and it would effect formulas). I’d work in the polar coordinate system. Instead of using formulas with x and y components, you will be using radius and angle components. Assuming your radius is constant we can get all of our formulas into angle components.

Another thing your going to need to understand is Rotational inertia. Its sort of like mass in your equations.(take this loosely, they are two completely different entities…just happen to be placed the same in formulas sometimes)

To summarize:
position=>angle
velocity=>angular velocity
acceleration=>angular acceleration
mass=>rotational inertia
force=>torque

Example:
Force=massacceleration
becomes
Torque=rotational inertia
angular acceleration

Now, I’ve covered all the basics you should need to understand the steps to find what you want.

First step is to find your rotational inertia. This can be a little confusing to do and I really can’t tell you how to do it without knowing the geometry of your kicker. I’d presume that many kickers can be solved for using a rod rotating about end, a ball, and the parallel axis theorem. This is going to get messy pretty fast and you might have to look up some videos to understand rotational inertia and why my math works.

I’m assuming your kicker consists of a stick with a weight at the end. I’m sure your design isn’t this simple, but with physics we choose to simplify things to make the math possible. The formula for the stick is simple.

Use: I=(mL^2)/3
m is the mass of the rod, include anything that is distributed between the axis and the end. L is the length of this rod.

Now, for the weight at the end we will assume it to be a ball. However, the formula is complicated because we are not rotating the ball about its center. Lets find the inertia due to its shape.

For a ball about its center use: I=(2mr^2)/5
m is the weight of the ball, include anything that is only at the end and wasn’t included in the rod. r is the radius, your really going to have to approximate this one.(it won’t be too big of a deal on your final answer, but it is good to be close)

Now for the parallel axis theorem. This theorem explains that a mass rotated about its center will have less inertia then the same mass rotated about a different axis. This formula helps us to calculate the additional inertia.

For the additional inertia use: I=m*d^2
m is the mass of the ball because that is what has been displaced. d is the distance of displacement. In this case d should be the L used in the above formula for the rod.

Now, to find the total rotational inertia just add the three inertias we calculated above and you will have your final inertia.

My main confusion is on what your wanting to find. You could find the torque required to get certain speeds of rotation. You could find the energy within the kicker which might relate to the distance you get. You could find the angular momentum which could also be used to find distance. Please explain the end goal a bit clearer and send me a PM so I’ll be sure to see it. I’ll gladly come back and go in more depth on parts I might have overlooked or to continue. Also, if you are having trouble putting in your data into the above formulas, just give me the info and I’ll help you figure it out.

Sorry I couldn’t finish the problem,
Jason

$@#$@#$@#$@#, all that math went straight over my head. So much for having a good grasp of BC Calc.

*edit Hum a language filter.

Here’s one for you -
The theoretical maximum speed the ball could leave the robot is twice the speed that the kicker strikes it at.

Example: Throw a tennis ball at a wall. It hits the wall, and leaves with the same speed that it came in with. Now imagine this is in space, and the wall has near infinite mass. Also, the tennis ball isn’t moving, but the wall is. Same scenario, and the same results.

A better way to go about this is trying to figure out how far and at what angle you want to kick the ball. After having those constraints, you can then figure out the velocity at which the kicker needs to have when it strikes the ball. Then, you can figure out the geometries and forces involved.

Oh, and then you can multiply everything by PI for good measure. Don’t ask me why, ask yourself why not. :stuck_out_tongue:

YOUR grasp of physics and mathematics appears to be top notch.
YOU’RE going to need to work on YOUR grammar though. :slight_smile:

Our kicker is based on pendulum theory.

But we didn’t calculate the force to kick the ball.
We worked with Angle(A) between Kicker and Ball and Tangencial Velocity(Vt) of the Pendulum.

So we used the follows equations:

Ymax (Maximum Height): (Vtsen(A))²/2g
Xmax (Maximum Distance): (Vt²
sen(2A))/g

You’ll find your maximum distance and maximum height based in Angle and Tangencial Velocity.

PS: If you wanna know the “perfect” kick (no considerations of air drag), in order that the ball doesn’t hit in your bumper and pass trough 2 BUMPS, you’ll need find the coeficients a and b of the equation F(x)=ax²+bx+c (the coeficient c you don’t need to find because your parable doesn’t cross the Y axis.

Man, I really don’t know if you could understand what I’m thinking about, because I never had discussed math or physics in english language, but this method worked for our team: http://www.youtube.com/watch?v=QMqMi_OIuRY
We vary the angle between the tests, and, as you can see, the ball pass trough Bumper and both bumps without hit in the bumper!

I apologize if my previous post seemed a bit bizarre. The math and physics I was trying to explain isn’t as bad as it looks(so don’t be scared of higher math on my part) but it probably looks foreign to someone that isn’t used to looking at things in that way. I guess that’s what I get for helping friends with their college physics, I sometimes go overboard on my explanations even though they are all well-intentioned. I would agree with the person that suggested that it is a testing problem rather then a physics problem to an extent. They both have their realm. Sometimes you need a really accurate answer and physics simply won’t give it to you because of all the variables that nobody figures for on the whiteboard. However, what if you don’t have time to test enough to tell which is best? Or can only buy parts for one design? This is when theory comes into play. It allows us to start when we can’t really start any other way.

I will never downplay the power of proper testing, however I often tire of people downplaying the power of theory. Neither is all powerful and neither will ever be complete without the other. The only way to be certain is to prove it with math and testing.