Engineering Challenge- Spring Loaded Kicker Edition
 

Working through a possible issue with our practice bot, I thought it made an interesting engineering challenge.

So you are at competition with your spring loaded kicker.
You have a motor that winds the kicker to the motor/gearbox stall point (motor stops before mechanical limits) with a quick release mechanism that allows the kicker to kick. You kick several shots and notice that the it would be good to kick the ball a little further (about 10%). All you can do is add or subtract spring rate. Should you:

A. More spring rate = more betterer!
B. Less rate is more when it comes to springs!
C. Nothing, no matter what you do it will stay the same.
D. Ask the pit admin for some of those frictionless bearings you've heard so much about.
E. Blame the programmer.

Please do the math:

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

I choose F (The Big Orange Hammer) because thats the way we do things on the killer bees (although my second choice would be E).

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Clarification: when you say spring rate, you're referring to the spring constant, right? In other words, bigger spring rate means bigger value of k? In that case. . .

If the motor cannot fully compress the spring, then getting a bigger, harder spring is not going to allow the motor to store more potential energy. In that case the limiting factor is the motor, so just modifying the spring's strength won't do anything for you. So, answer C it is.

Right?

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

big orange hammer? 862 has one of those...we call it Yippie Skippie
but in all seroiusness of the thread i guess A
see you @ Troy 33!

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Try again. Highlight below for a hint. As IKE said, please do the math. If you show your calculations we may be able to help you see your mistakes.

The maximum force is the constant in this set of equations

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Yes (to the first part), Spring Rate is referring to K or Spring constant (force/unit distance... N/m or lbs/in or stones/cubit...).

As far as the silver text goes... your may be thinking in the right direction, but some more math would help.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

My guess is c.
The motor can give the spring potential energy, and changing the length or rate of the spring will not change the potential energy the motor is giving away.
With the same potential energy, the spring will always kick with the same speed (energy calculations).

I hope I'm right.. never had any "challenges" like this :confused:

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Bryan, you are wrong. It has nothing to do with my code. I blame the mechanical team.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

This was inspired by a Thundershicken that headed South for the Winter which was inspired by an amatuer robot guy that turned pro that was inspired by...

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Decrease spring rate and preload the springs to produce a more constant spring force.
When graphing Force vs displacement, without preloading you get a triangle. Preloading the springs makes it a trapazoid or in an ideal world a rectangle. The area under the curve represents Work.
More work = more kick

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Analyzing the problem:

P=MV You want full travel out of your system to give you the most time for your mass to accelerate and have the largest velocity and therefore momentum (P) at impact.

Analyzing the choices:

A. More springs

Cause your motor to stall with less displacement on the spring. Most likely an equal force, but there is less distance between your kicker and the ball for the mass of your kicker to accelerate.

B. Less springs

More stroke -> more time for the mass to accelerate -> more momentum at impact -> farther shot

C. Do nothing

What kind of engineering is this?

D. Frictionless bearings

Those'd help but I don't think the pit admin table has any.

E. Blame the programmer

As much as I'd like to....

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Dang, I knew I should have gone with answer E.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Assuming the kicker is mechanically limited in the forward (kicked) position, ADZDEBLICK has it exactly right. Decrease the spring rate, and pre-load the spring just enough so that the motor will stall before it hits a mechanical limit in the reverse (armed) direction.


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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

G: Not enough design parameters are given. More information is needed.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

You're making an important assumption about the design of the kicker here, which may not be true. If the kicker is set up like a big pendulum, with significant mass and significant travel time before contacting the ball, then I think you're correct.

What if that's not the case? What if your kicker is designed to work without significant mass and without any travel time before contacting the ball? In other words, you have a relatively mass-less kicker already in contact with the ball when you start to accelerate it. That's essentially our design, and it works great (clears 1 bump easily, scores very well). Think of a slingshot, and you'll have a good idea of how the physics of it works.


Now, everyone should keep in mind that a robot cannot create energy. The form of the energy can change, but it can't be created.

In this situation, a motor is able to exert a certain amount of torque (which can be translated into a linear force very easily) before it stalls. In short, this means it can pull back the kicker a distance x such that the linear force the motor is outputting is equal to kx. The stall torque for any given motor is a constant.

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.

So, does increased draw help in this situation? Well, that all depends on the design of the kicker :)


We ran into this problem early in the season. With our design, increasing the draw really doesn't get you anything. Adding a second motor and more springs, however... :)

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

It doesn't matter.


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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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

... but the answer is still correct.

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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.


~

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Trapazoids kick farther than triangles

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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


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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.


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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

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.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

The approach we used with our triangular pendulum kicker was based on the fact that a professional soccer players only remains in contact with the ball for ~1/100th of a second. So, our goal was have our kicker rapidly reach a high velocity BEFORE contacting the ball. The only way to accomplish this was to sustain a high force THROUGHOUT THE TRAVEL of the kicker, BOTH before AND during contact with the ball.
To produce this result, we preloaded our bungie set so that there was at least 20 lbs force at full extension. We then went for about 10" of winder retraction of the kicker to build up to a force level of between 80-100 lbs. Having a kicker mass at about 1.5 times the ball's, this produced some serious acceleration of the kicker and very high energy transfer during kicks.

FROM AN EMAIL SENT TO MY TEAM EARLIER IN YEAR ON A FANTASTIC WEB BOOK:
This web document has the most in depth math/physics analysis of ABSOLUTELY EVERYTHING relating to soccer!!
=> http://www.scribd.com/doc/6726997/ScienceofSoccer

Their analysis of kick, bounce and spin are useful, and worth reviewing briefly.
Some items covered include:
1) good kicks go 70 miles per hour & contact with foot lasts ONLY 1/100 of a second
2) ball spin of around 10 RPS can accelerate a ball on landing by 5-10 miles per hour

Based on my read of this info, I am starting to think that kicks traveling the length of the field will not be so easy to accomplish.
We likely need to have a minimum 3lb mass traveling at ~30+ MPH impacting the ball to have it travel the length of the field, clearing both bumps.
Remember that we also need to instantly absorb the momentun impact of this 3LB. mass against our robot frame too.


-- Dick Ledford

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

I find it funny that no one started using actual math until halfway down the second page of the thread......

It's all because of the x^2 value in 1/2Kx^2 which is the energy stored in a spring which in a lossless world translates into kinetic energy of the ball. x^2 will increase the energy much faster than the K will decrease it in any situation. Messing with the equation and substituting in the equation for force of a spring defined by Fs=Kx, we get Pe=1/2(Fs/x)x^2 which = 1/2Fs(x) or in other words, at a constant Fs (stalling a motor) the potential energy stored in a spring is directly proportional to how far you pull it back (x).

Also, stalling motors regularly = bad.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

This is very easy to see graphically. See the first attached GIF. The area under the curve of force vs distance equals stored energy. The area under the curve of the Weak Spring (shown in red) is larger than the area under the curve of the Strong Spring (shown in blue). Both the Weak and Strong spring areas have the same height, but the base of the Weak spring area is longer, so the area is proportionally greater (area of a triangle is 1/2 the height times the base).

Your analysis completely ignores pre-loading of the spring though. Most teams using spring-assisted kickers pre-load them.

By using a greater pre-load on a weaker spring, you can get more usable kicker energy at the same stall force and same distance. See the second attached GIF.



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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Very true. What would be even better, using this logic, is a constant-force spring. For a given usable length they are much smaller than a coil spring of the same rate. Check out McMaster.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

If the motor stalls before the physical limit, you could:
-add a reduction to produce more torque
-make sure that the motor is applying force parallel to the kicker to the path of the kicker (or perpendicular if it works as a pendulum)

If the physical limit is the issue:
-increase spring constant
-increase distance of tension/compression
U=.5kx^2
because the motor would still be able to apply enough force to pull it back further, the spring constant/distance could be increased and therefore the spring potential would increase as well.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

The source of the ultimate amount of energy input to the spring is from the motor that is compressing or stretching the spring. The problem was stated that the motor loading the spring stalled. Bad for the motor but what is even worse the power or energy out from the motor is Zero. Max motor power output is usually when the speed of the motor is loaded down to about 1/3 of max RPM as I remember it. Therefore the trick to get the max power available from the motor into the spring is to keep the motor from stalling which means a smaller spring constant. You could take it to an extreme in a frictionless vacuum environment but we are dealing with the real world here so get a “weaker” spring and plan on greater travel for loading it. Good luck!

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

The power is zero but the torque is maximum.

The maximum power output occurs at 1/2 maximum rpm.

This thread is not about getting maximum power from the motor. It is about storing the maximum energy in the spring. They are two different problems.

a weaker spring with greater travel at the same stall force would indeed produce more stored energy in the spring.

but greater travel is not necessary to get greater stored energy: a weaker spring can be used with greater pre-load to stall the motor at the same travel, and this will produce greater stored energy in the spring.

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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

You are right that max power is at ½ max RPM. I was thinking mistakenly about max efficiency occurring in the lower half of max RPM.

The question IKE posed is what to do to increase the distance that the ball is kicked with the constraint of only adjusting the “spring rate” ie spring constant either increasing or decreasing it.

To kick further you need more energy stored in the kicking energy storage mechanism i.e. the spring. The motor is the only method available to add the extra energy required. I assume that this is not the first kick which can have lots of stored energy due to preloading the spring in the pits but rather a subsequent kick relying only on the motor to reload the spring with energy. The present condition is that the motor is stalled i.e. angular speed is zero. Even though the motor is producing max torque, the power out of a stalled motor is zero because angular speed is zero (Power = torque x angular speed), and the energy output is also zero because angular speed is zero ( K = ½ x Moment of inertia x (angular speed)sqruared ). The trick therefore is to not allow the motor to stall but continue to rotate by lessening the force i.e. torque on the motor. That is accomplished by lowering the spring constant.

Energy stored in a spring is the area under the F(x) by x function. If the force is limited to Fmax (stalled motor) the only way to increase stored energy is to extend the distance X (amount spring is compressed or stretched) which is accomplished by decreasing the spring constant. By decreasing the spring constant 10% the amount of stored energy goes up 9%.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Since the stall torque of the motor is constant, and that the potential energy stored in the spring is proportional to x^2, the motor could be moved such that the axis of rotation is closer to the piece you are pulling back.

At the stall point the lever (I'm assuming that lever attached to the motor / gearbox is being used to pull back something connected to the spring) is in equilibrium. The torque from the motor perfectly balances the moment caused by the force from the spring.

The moment caused by the force on the spring is equal to F*d where d is the distance from the point of rotation, by moving the motor closer to the spring (assuming you have the time to do this and it is feasible for your design) the spring could travel further before causing the motor to stall. This would increase the potential energy stored in the spring.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Thanks for all the great inputs. Especially those that went into more detail than I think, or I have a hunch.

Yes, this was assuming no other conditions but rate were changes, but the tangents were good too.

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

"Pre-loading" the spring, in the context in which that phrase was used in earlier posts in this thread, means that the spring force is not zero at the end of the kick. For example, suppose the "spring" is four 20" parallel pieces of latex tubing. Reduce the spring rate by getting rid of 2 pieces, and cut the remaining 2 pieces down to 12" to pre-load them. Please look at the graphs in earlier posts.

No, extending the distance is not the only way to get more stored energy. You can get more stored energy with the same distance by using a weaker spring and pre-loading it (see definition of "pre-load" above) to a higher value. This is illustrated graphically in earlier posts.



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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Hi Ike and everyone,

Excellent, excellent discussion. Many of your conclusions match what I got when I took F = ma = -kx, turned it into a = -kx/m and found x(t) and v(t) using hypothetical values of m and k. If all we care is the velocity of the kicker and momentum at the end of the swing, having smaller spring rate and longer draw do indeed create a higher velocity and momentum in the end. Same with increasing the kicker's mass (it reaches a peak after a while).

So, I started looking for the actual spring rate of springs 115 used on the robot this year, and ended up digging around the spring manufacturer's site as well as through Shigley's Mechanical Engineering Design textbook. I found a nice pdf in Century Spring's website (our springs' manufacturer) that talks about spring rate and various other considerations: http://www.centuryspring.com/pdfs/230-289.pdf, and I found the mechanical property of Hard-drawn wire from Shigley's. I obtained a theoretical spring rate that matches closely with what's listed in the catalog specs (We used 0.563 in OD, 8.5 in length, and 0.054 in wire diameter springs).

Then I took one of our springs home, hang it up with a coat hanger, tried to put weight on it, and tried to calculate the actual spring rate. To my surprise, the rate is different at different load (5 lbs, 10 lbs, 15 lbs). It's much higher at lower load and approaches the theoretical value as load increases (the highest suggested load for this spring is 9.2 lbs).

Then I pulled up a spreadsheet I made about a week ago when I was messing around with kicker/cable/pneumatic geometry, and tried to calculate effective cable tensions at different kicker angle.

Long story short, I will be posting a white paper on spring loaded, pneumatic drawn kicker systems in the near future.

-Ken L

Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Under no circumstances will increasing the kicker's mass (while holding all other parameters constant) result in higher velocity of the kicker at the end of the swing.


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Re: Engineering Challenge- Spring Loaded Kicker Edition
 

Right, I looked at my wording again, and mis-spoke if you read it to say it would result in a higher velocity of the kicker. What I meant to say is increasing the mass will also increase the momentum of the kicker, but only to a point as it approach a certain value.

In my mind, higher momentum coming out of the kicker mean a higher velocity on the soccer, but it was getting late and I wasn't careful about my wording.

Oops. Getting too old for late nights I suppose.