How hard will a ball deflection hit?

[Culvan Van Li]

So my question is how much force should a robot mechanism be able to withstand to survive contact with the ball? I'm leaning towards 25lbs, but I wanted to solicit feedback since this isn't my strong suit. Has anyone got a good number for how much force the ball is likely to impart when it bounces off of a robot? Does anyone have a force plate they could drop a ball on to get us some numbers? We could also get some good calculations from high speed camera footage of the ball bouncing.

Here's the current model I'm working on:
A ball launched at a 45 degree angle rises to a height 2 meters above it's launch point and falls the same distance before it impacts the robot of interest. Energy at the time of impact can be calculated as Mass*Height*Gravity/(Sin Angle) = 1.136*2*9.81/(sin 45) = 31.53 joules. From there I can work back to figure out the incoming Velocity = sqrt( 2 * Ke/mass) = 7.449 m/s (downward). Observing the ball as it bounces, our ball seemed to bounce about 50% of the height is was dropped from utilizing the equations above I come up with a departing velocity of 3.97m/s (upward). The total change in velocity is 11.42 m/s. Force = Mass * Acceleration and Acceleration = Change in Velocity / Change in Time. Unfortunately I don't have the ability to measure how long the ball is in contact with the ground. I guessed at a quarter of a second (does anyone have a better number?) That gives us an average force of 51.9N or 11.7lbs.

I know I'm not taking air resistance into this equations and many other smaller factors, but I think the numbers I have listed above will dominate the equations. I also think that it is entirely possible to have balls thrown harder than I've listed. Additionally the ball will act like a spring and the force it exerts will be defined by the equation F= K*x where K is the spring constant and x is the amount it has been deformed. This implies there will be a linear variation to the force as it bounces. That probably means that maximum force would probably be double the average. That implies that the max force asserted based on my model is over 23 lbs.

Andy

[Alan Anderson]

You might find the equation Ft = mv useful. The amount of momentum transferred to the ball is equal to the product of the force used to impel it multiplied by the amount of time that force is applied.

If you shoot the ball quickly (not quite the same as "shoot the ball fast"), you will have greater forces involved.

[Culvan Van Li]

I've got an update to our situation. We made a prototype out of plywood and 2x4s. We tossed the ball at it a few times. On the fourth impact, it split a 2x4 along the grain. Considering how strong the build looked I'm surprised by the damage. I think the lesson we learned was that there is more force imparted by a strong ball hit than you would probably expect. I'm glad this was a discovery in week 1 rather than week 3 or later.

Originally Posted by Alan Anderson

You might find the equation Ft = mv useful. The amount of momentum transferred to the ball is equal to the product of the force used to impel it multiplied by the amount of time that force is applied.

It is interesting how similar that equation is to F = MA = M * Change in Velocity / Change in Time. It should be possible to calculate resulting vectors in a collision based on using the conservation of momentum and the conservation of energy.

Andy

[Mike Schreiber]

Originally Posted by Culvan Van Li

It is interesting how similar that equation is to F = MA = M * Change in Velocity / Change in Time. It should be possible to calculate resulting vectors in a collision based on using the conservation of momentum and the conservation of energy.

Andy

Good observation. The momentum-impulse equation is the integral of Newton's Second Law. Integrating force over time gives you F*t, and the integral of acceleration is velocity.

[RSchorr]

I have been running the force, momentum acceleration equations and having to guess on the time duration for computing deceleration. I was going to state that my initial calculations led me to expect a force of 50 or 60 lbs rather than the 25 lbs you started with. However, your statement about the 2X4 splitting along to grain makes me wonder what you were getting as a force (force of impact).

Any thoughts on this?

[blaze8902]

Well, 2x4s are anisotropic, so orientation of the grain is important, as is the direction of the force applied. I'd say either your wood had an un-surreptitiously placed grain, or your team members have got strong arms.
How long was your 2x4, and where was it supported? a 4' 2x4, suspended on both sides will break at about 317lbs, assuming you're breaking it by applying the force on the 4" side, where it will be easier. If you threw it at the 2" side, it's more like 740 lbs. These calculations were done based on the assumption that the grain of the wood ran lengthwise down the length, as this is the most common. If your grain was another way, then it'd be easier to break. If that's the case, you should consider finding a new wood supplier, because that's both unusual and bad for construction.

Edit: The length, position, grain orientation, and support location are all factors. If you would like, I can calculate the force required to break if you supply me with this information. (I'm no expert on wood, so I can't say for sure just how accurate I can be if you say that your grain was not oriented lengthwise.)