Engineering Challenge- Spring Loaded Kicker Edition

[Ether]

Originally Posted by ICanCountTo19

I was making the assumption that this problem was based off of the Killer Bee's robot which does have a distance between the kicker and the ball in which the mass accelerates.

It doesn't matter.


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[theprgramerdude]

I would go with D; it would mean less momentum and energy loss due to friction on the way down. Assuming! that the kicker was put together by a bunch of monkeys (I'm a programmer), it might just have too much friction loss to get 100% effiiciency from the spring.

[IKE]

Originally Posted by theprgramerdude

I would go with D; it would mean less momentum and energy loss due to friction on the way down. Assuming! that the kicker was put together by a bunch of monkeys (I'm a programmer), it might just have too much friction loss to get 100% effiiciency from the spring.

If they are highly trained monkeys does this make a difference?

[Ether]

So, what do we do with k? Well, if we increase k, we get a shorter draw but the same amount of force. If we decrease k, we get a longer draw, but the same amount of force. In the end, the force (remember F=ma?) remains the same (as does mass and acceleration). The only thing we change is the draw.

You are forgetting about pre-load.

1) Assuming that both the strong spring and the weak spring have the same pre-load force, then the weak spring will be pulled back further and have more stored energy.

or

2) Assuming you increase the pre-load on the weaker spring just enough that the motor stalls at the same position as the strong spring, you will also have more stored energy in the weaker spring.

If you have trouble believing this, make a simple graph of force versus distance for each case and look at the area under the curve.


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[JamesCH95]

In short: make the kicker lighter, it will more efficiently transfer momentum to the soccer ball

I like to look at these sorts of problems holistically. Your motor can only put out a certain force and your kicker can only retract a certain distance, so you have a maximum amount of energy you can store in the springs. EDIT: Lowering the spring rate *might* help if you can work out all the details, i.e. fit a spring on that will get you a greater amount of stored energy, K*x^2. Constant-force springs would be an interesting option to explore, they would allow you to maximize the energy extracted from the motor by keeping it under a higher constant load, rather than a progressive load as shown in Ether's plot.

When considering elastic impacts I found through momentum calculations that for the same amount of applied energy (Force*distance or torque*swept angle) it was beneficial to decrease the mass/inertia of the kicker to improve kicking distance. If you need a 10% increase in range you might need to decrease your kicker's inertia by ~10% but I cannot give you exact numbers without knowing details about your kicker.

This is a good simple paper explaining basic physics behind kicking a soccer ball:

http://www.serioussoccer.net/Documen...csofSoccer.pdf

[AustinSchuh]

Originally Posted by JamesCH95

In short: make the kicker lighter, it will more efficiently transfer momentum to the soccer ball

Curious. Experimentally, we found that adding mass to one of our prototype kickers increased it's range from something like 5 feet to 15 feet. Then again, that prototype kicker was an extremely light and small piece of metal.

If you run some numbers with Va = 0 and Vb = 1, and the coefficient of restitution equal to 1 in the equation given at Wikipedia, you will find that equal masses of the two objects will result in perfect momentum transfer. I'm going to go out on a limb and guess that since our kickers are swinging around an axis, what you really want is equal moments of inertia around the axis for perfect energy transfer.

[Ether]

Originally Posted by AustinSchuh

If you run some numbers with Va = 0 and Vb = 1, and the coefficient of restitution equal to 1

The COR is nowhere near equal to 1. So your conclusion is not valid.


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[IKE]

Mass of a kicker is important to a point. the kicker and the ball need to be in the same order of magnitude otherwise there are different governing equations.

Think of it in terms of tiny mass or really big masses. The ball itself weighs about 0.5kg or roughly 1.25 lbs. If your kicker is really light. Like 1.25 OZ. then in order to impart the kinetic energy needed on the ball it will need to go at sqrt(16) or 4 times the speed. This is really fast. If we go a step further, then say about 1/16th of an ounce, then it would need to be another 4x of 16x faster. For something this light, it would have to go really fast and might puncture the ball. Thus more surface area would be required, and it would loose a lot to areo losses before striking the ball.

Now go the other extreme. Say you use a 16 lbs sledge hammer. From an energy standpoint, you would only need to go 1/4 the speed for the ball, but in this senraio you would likely just push the ball instead and not get a big kick out of it.

Too little and it is difficult to store and transmit enough kinetic energy. Too massive and it is difficult to get the mass up to an effective speed.

[JamesCH95]

Originally Posted by AustinSchuh

Curious. Experimentally, we found that adding mass to one of our prototype kickers increased it's range from something like 5 feet to 15 feet. Then again, that prototype kicker was an extremely light and small piece of metal.

If you run some numbers with Va = 0 and Vb = 1, and the coefficient of restitution equal to 1 in the equation given at Wikipedia, you will find that equal masses of the two objects will result in perfect momentum transfer. I'm going to go out on a limb and guess that since our kickers are swinging around an axis, what you really want is equal moments of inertia around the axis for perfect energy transfer.

The CR of a soccer ball is in the region of 0.6-0.7, so try your calculations again.

A large contact area on the ball is also important for good momentum transfer, large deformations of the ball result in lost energy, so a big striking surface is probably better than a small one, in general. I would imagine your small thin piece of metal was not very stiff, again a very inefficient way to transfer momentum because the kicker will deform and absorb energy. Things designed to transfer momentum like pool balls and sledge hammers are all very stiff.

Originally Posted by IKE

Mass of a kicker is important to a point. the kicker and the ball need to be in the same order of magnitude otherwise there are different governing equations.

[...]

For something this light, it would have to go really fast and might puncture the ball. Thus more surface area would be required, and it would loose a lot to areo losses before striking the ball.

[...]

Too little and it is difficult to store and transmit enough kinetic energy. Too massive and it is difficult to get the mass up to an effective speed.

Your first point is very valid, the mass of the kicker needs to be compromise between efficiently absorbing the energy that's applied to it and transferring that energy to the ball. Though for spring-fired kickers lighter is better because springs can deliver their stored energy at very high power levels.

Any kicker of reasonable design (i.e. not a giant flat board) would have to be traveling incredibly fast to see any sort of significant aerodynamic losses, I would consider them negligible for any design.

Final point is again correct, but I feel that as long as the kicker is stiff enough (ours is 4130 steel tubing) then faster is better. IIRC the PGA has banned the use of certain clubs that are super-light and have extra-hard striking faces because they can drive a golf ball significantly further than a softer, heavier club.

[AustinSchuh]

Assuming a COR != 1, we get that the kicker should weigh M_ball / COR. What ever initial conditions exist, it is trivial to find a mass of the kicker that will result in the kicker transferring all it's energy to the ball. And for most sane initial conditions, you will want the masses to be within a factor of 2 of each other.

I've been thinking about this for a while, and this post finally got me to sit down and derive what happens when the kicker is spinning and hits a ball. Something bugged me about using equations derived for linear impacts when modeling something that is swinging.

Lets start with the assumption that angular momentum is conserved.

I_a w_a + I_b w_b = I_a w'_a + I_b w'_b

Then define an "angular coefficient of restitution" that's similar to the coefficient of restitution for linear collisions.

-w'_a + w'_b = COR (w_a - w_b)

Solving the linear system of equations gets us that

w'_a = (I_a w_a + I_b w_b + I_b COR (w_b - w_a)) / (I_a + I_b)

w'_b = (I_b w_b + I_a w_a + I_a COR (w_a - w_b)) / (I_a + I_b)

So, when a rotating kicker is hitting something, we want the moments of inertia to be similar. Or, if the COR isn't one, then the moment of inertia of the kicker should be I_ball / COR, where I_ball is the moment of inertia of the ball around the axis that the kicker spins around. Very similar to the original billiard ball case.

Our small piece of metal was the end of one of the old IFI frames that looked like a U. So, it was a ~1/2 lb quite stiff piece of aluminum that had about a 4" radius. It's moment of inertia around the kicker axis would have been quite a bit lower than the ball's around the kicker axle, resulting in a shortened kick distance because the kicker would bounce off the ball instead of kicking it. Which is consistent with what we observed.

[Vikesrock]

Alright, I've seen enough answers that I think are wrong to make begin to wonder if it is me that's wrong so here's what I got:

To get the farthest kick we want to impart the maximum kinetic energy on the ball. Assuming all the potential energy in the spring is converted to kinetic energy when firing, this means we want to maximize the potential energy stored in the spring.

F=kx
E = 1/2*kx*x

F in this case is the max motor force which is constant, therefore kx is constant. If kx is constant then E is proportional to x, therefore we want to maximize x.

Going back to the first equation we now know F (the max force the motor can apply to the spring based on gearing and lever arm) and X (the max spring extension before reaching the mechanical limits) so we can solve for k.

For x to increase, k will have to decrease so my answer is B, less spring rate.

[IKE]

Ok, in case this is stumping anyone this weekend, I will put my thoughts down in White:
So with the spring loaded kickers, most need a mechanical endstop to avoid penalty. Because of this, the ball will tend to follow a given trajectory. If you use trajectory physics, you can calculate the given distance if you know a couple parameters for this problem, ball speed and angle would give you distance. How? and How is velocity related to the distance? Well the lateral distance covered will be equal to the time in flight multiplied by the lateral component of the ball velocity. The time in flight is dictated by the vertical component of the ball velocity. Since gravity only slows the vertical component, 1/2*t=Vy/g with g being the gravity constant. Therefore the distance (d) traveled is d=Vx*t=Vx*2*Vy/g. This means that really distance is proportional to a constant (c) multiplied by the square of the velocity (since Vx and Vy have a constant relationship). Thus d is proportions to V^2.
So how do we find V? Well 1/2mV^2 is the kinetic energy imparted on the ball by the kicker. Where does this energy come from? The spring. Thus the spring energy 1/2K*x^2 is directly proportional to the ball kinetic energy 1/2*mV^2. From this we realise we want to maximize the kinetic energy.
Now for this particular problem, as many have noticed, the motor stalls thus it is torque limited thus Kx (spring force) is maximized. As we have seen though, energy is really what we are looking for. lets assume both K = 2 and X = 1 Then the energy stored is 1/2K*x^2=1/2*2*1^2=1 Now if we reduce the spring rate and hold the force constant Kx so that K = 1 and X = 2 then 1/2*1*2^2=4/2=2. We actually doubled the energy going into the kicker. This assumes though that you have the extra travel. what if you don't have the extra travel? Can weaker springs still give you more energy? Initially you would say no, if they both pull the same distance, then the stronger spring should store more energy. This assumes no preload though! so back to our earlier example. K=2 and X=1 gives us an energy of 1. Now with k =1, X can equal 2, but there isn't enough room. therefore we pre-stretch the spring so that it ends up with a total stretch of 2. Thus the energy for that is E= 1/2*K*(x2)^2-1/2*k*(x1)^2 where x2=2 and x1=1 thus E=1/2*k*(x2^2-x1^2) or E= 1/2*1*(2^2-1^2)=3/2 thus even in a restricted space, we still have 50% more energy with the weaker spring rate.

So my answer is B sometimes with springs, less rate is more.
Good luck to the teams competing this weekend.

[Ether]

Originally Posted by IKE

Ok, in case this is stumping anyone this weekend, I will put my thoughts down in White:

It's much simpler to see if you just plot a simple graph of force versus distance, and look at the area under the curve. See attached GIF file. The area under the curve is the stored energy. The area under the red curve is far larger than the area under the blue curve.


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[ADZDEBLICK]

Originally Posted by Ether

It's much simpler to see if you just plot a simple graph of force versus distance, and look at the area under the curve. See attached GIF file. The area under the curve is the stored energy. The area under the red curve is far larger than the area under the blue curve.


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Trapazoids kick farther than triangles

[Ether]

Originally Posted by Vikesrock

Assuming all the potential energy in the spring is converted to kinetic energy when firing

If the spring is pre-loaded, this is not true...

my answer is B, less spring rate.

... but the answer is still correct.

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