How does the placement of sensors affect stress in a squeeze ball?

[Anuroop]

Hi,

I have two squeeze balls with pressure sensors inside them. One squeeze ball(Ball A) has a sensor placed at the center while the other one(Ball B) has two sensors, placed very close to the two opposite surfaces. These squeeze balls act as an input to a game developed. Depending upon the squeeze force level, characters in the game will move.

My next aim is to theoretically prove that while squeezing the ball,maximum stress is acting on the sensors placed in Ball B as it is closer to the surface. In a way I am trying to prove that the input force exerted by the user is applied to the game efficiently when we use Ball B.

So how do I prove it by writing few equations? Is it just calculating the radius of contact area & then using it to find out the stress acting?

Regards,
Anuroop

 

[CWatters]

I might not be the best person to answer but...

If you treat the ball as a liquid being compressed uniformly I would expect the pressure to be roughly the same everywhere inside it. If it's not then I suspect its due to the non-uniformity of the set up?

 

[Anuroop]

I might not be the best person to answer but...

If you treat the ball as a liquid being compressed uniformly I would expect the pressure to be roughly the same everywhere inside it. If it's not then I suspect its due to the non-uniformity of the set up?

Thanks for the reply. My explanation for stating that maximum stress is near the surface goes like this.

The pressure is a function of the force applied divided by the area it is spread over. And if you think about it, any cross section of the ball between the two points must have the same total force transmitted. But the area of that cross section has a maximum across the middle of the ball leading to the lowest pressure and it has a minimum area near the points of contact giving the highest pressure.

I am not 100% confident about it & I am looking for some theories/ equations or some experiments to prove it.

 

[Delta2]

When we press at 2 points a squeeze ball we essentially generate 2 pressure waves that travel with the speed of sound in the air inside the ball (i assume the ball is filled with air). It isn't so easy to find where the total pressure wave will have minimum and maximum but in a very short time (in a time scale of d/s where d the diamter of ball and s the speed of sound) the two waves will dissipate and the pressure will be about the same everywhere inside the ball. So i think if we want better more accurate reaction times comparable to d/s we should use ball B (which i doubt because d/s would be very small) otherwise ball A will give about the same results.

 

[jbriggs444]

If you model the ball as a fluid-filled, flexible skin then, as CWatters points out, pressure will be uniform throughout the interior. You get the illusion of locally greater pressure at your fingers, not because the fluid pressure is greater locally, but because the curvature of the skin is concave. The tension of the (concave) skin adds to the pressure of the fluid. By contrast, where you are not squeezing, the tension in the (convex) skin cancels the pressure of the fluid.

I suspect that the interior of a real squeeze ball can support sheer stress so that the model as a fluid-filled skin may not be correct.

 

[CWatters]

If the sensors are at two points on the surface of the ball I reckon they will be sensitive to how the ball is held, eg to the orientation of the ball in the hand. If the sensor is in the middle of the ball it should be immune to orientation.

I am trying to prove that the input force exerted by the user is applied to the game efficiently when we use Ball B.

Putting it in the middle might make the sensor less sensitive but you should be able to fix that.