View Single Post
  #1   Spotlight this post!  
Unread 19-02-2014, 18:04
IKE's Avatar
IKE IKE is online now
Not so Custom User Title
AKA: Isaac Rife
no team (N/A)
Team Role: Mechanical
 
Join Date: Jan 2008
Rookie Year: 2003
Location: Michigan
Posts: 2,149
IKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond reputeIKE has a reputation beyond repute
Cushioning Your Hard Stop

A little late, but something your team might want to plan during your unbag or practice day.

At Week 0 and looking over several catapults, I have observed 3 types of hard-stops for catapults that will likely lead to broken parts.

The frame stop for the catapult, where part of the catapult (usually near pivot) stops the frame of the catapult. Putting a little bit of hard rubber at this position will keep the frame from hammering itself to death.

Catch strap to frame. This uses a stiff rope, strap, or cable to stop the frame at a given position. This can put a lot of stress into the frame, pivot, or anchor point. Having a compliant bushing under the loop of the anchor or frame will help cushion the end travel here as well.

Last but certainly not least is using pneumatics as the driver/end stop. This is potentially dangerous as the extra inertia of the catapults and high speeds can cause the ends or pistons to get damaged on the pneumatic cylinders. I have heard some of them have an internal cushion to help keep from having these break the cylinders. I would highly recommend a catch strap for those catapults as well.

The physics behind this problem is similar to crash dynamics.

The spring or motorized element has a lot of power that translates into a whole bunch of kinetic energy (some in the ball, but some in the frame of the catapult.
So... KE=Kinetic energy = 1/2*M*V^2=1/2*K*x^2 (stored energy of the spring).

This helps explain why firing without a ball can be worse than with a ball. With the ball, a lot of the spring energy goes to the ball. When dry firing, all of that energy goes to the frame.

Energy though doesn't break the parts, but forces do. This is where the "cushion" comes in.
Energy also equals force time distance: or KE=F*D
So KE=1/2*M*V^2 of the frame = F*D
Where force is at the point of the stop, and distance is the deflection at that stop. As the stopping distance goes to 0, then the Force goes to infinity!

Well, not exactly. With forces approaching "infinity", all sorts of things start acting like springs. and compressing/stretching at the local level. When you get into your mechanics of material, it is known as strain. Strain them too much, and they stay permanently deflected. Permanently deflect something too many times, and it will break or stop functioning.

This is where the little block of stiff rubber comes in. Let's say that the mechanism has an inertia equavalent to about 1 kg, and is traveling at 13 m/s when it reaches end of travel. This is 1/2*1*13^2 or... about 85 J of energy. If your stopping distance was 1m, then the force would only be about 85 N (19 lbs). Not too bad. If the stopping distance was consistant, and 0.1 m, then the forces go up to 850 N (190 lbs). A little scary, but a lot of things can handle 200 lbs. Of course, 0.1 meters is about 4 inches which is a fair amount to stop something. Now, a lot of these catapults have ahrd stops. At 0.01m (1 cm or about 0.4 inches) is 8,500 N (1,190 lbs) starting to become a lot of load. 0.001m (0.040 inches) is a "hard stop", and would be 85,000 N. Since the ball weighs about 1 KG, removing the ball would double this set up. or 170,000 N (38,217 lbs). Even with 1" of cross section, a lot of aluminums will deform. The neat thing is, adding in 1/4" thick of stiff rubber will reduce these stresses immensely. For instance, this stuff: is set up for 2,500 psi. Thus, a square inch of it for two points of contact would match up reasonably well (0.040->0.25 reduces forces to 6,000 lbs.. one on each side would then be 3,000 psi). This would be some easy to apply stuff with adhesive backing. 70-90A would be my recommendation.