Engineering Challenge- Spring Loaded Kicker Edition

[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.

[RRLedford]

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

[bman]

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.

[Ether]

Originally Posted by bman

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).

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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[JamesCH95]

Originally Posted by Ether

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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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.

[KrazyCarl92]

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.

[jonboy]

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!

[Ether]

Originally Posted by jonboy

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.

The power is zero but the torque is maximum.

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.

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

Therefore the trick to get the max power available from the motor

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.

get a “weaker” spring and plan on greater travel for loading it.

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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[jonboy]

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%.

[Evert Timberg]

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.

[IKE]

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.

[Ether]

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.

"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.

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.

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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[Ken Leung]

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

[Ether]

Originally Posted by Ken Leung

...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...

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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[Ken Leung]

Originally Posted by Ether

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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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.