Engineering Challenge- Spring Loaded Kicker Edition

[IKE]

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:

[BJC]

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).

[JackG]

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?

[Wildcat]

Originally Posted by BJC

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).

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!

[Vikesrock]

Originally Posted by JackG

Right?

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

[IKE]

Originally Posted by JackG

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. . .


Right?

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.

[Mori1578]

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

[apalrd]

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

[IKE]

Originally Posted by Mori1578

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

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...

[ADZDEBLICK]

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

[Mike Schreiber]

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....

[JackG]

Originally Posted by Vikesrock

Try again.

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

[Ether]

Originally Posted by ADZDEBLICK

Decrease spring rate and preload the springs to produce a more constant spring force.

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.


~

[sanddrag]

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

[Jon Stratis]

Originally Posted by ICanCountTo19

B. Less springs

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

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...