Vex Note: How a single flywheel ball shooter minimizes the effect of ball mass variations

May 28, 2015

Nothing but Net 2015/2016 competition game involves shooting 4 inch balls that can have a 10% variation in mass.    We know that trajectory range ,R = V^2/g*sin(2*theta) so it  is dependent upon the square of the ball release speed , V, and shooter elevation, theta.   Mass does not enter into the equation unless it affects V.

Ball release energy :

Suppose we use a Vex 5″ diameter wheel as a flywheel and rotate it a w_wheel angular speed.      As the ball leaves the shooter, it will have a V = r_wheel*w_wheel/2.  e.g. half of the flywheel tangential speed.    The ball will have a spin rate , w_ball = V/r_ball.    The energy of the ball, E_b , is the sum of the ball translational energy and rotational energy.

E_b = 1/2*m_ball*V^2 + 1/2*I_ball*w_ball^2

where I_ball = 2/5*m_ball*r_ball^2 (solid sphere of uniform density).

so Eb =  1/2*m_ball*V^2( 1+2/5)  .   (corrected 5/29 Was 1/2*m_ball*V^2( 1+4/5)  So the rotational energy adds  40% more to the translational energy.  Rewriting in terms of w_ball gives

E_b = .7*m_ball*w_ball^2*r_ball^2  

Wheel Energy:

E_wheel = .5*I_wheel*w_wheel^2.  where

I_wheel = m_wheel*(r_wheel*.84)^2  (ref blog post https://vamfun.wordpress.com/2015/05/17/finding-the-moment-of-inertia-of-a-vex-wheel-using-parallel-axis-theorem/)

Energy Conservation:

E_wheel_initial = E_wheel_final + E_ball     This assumes that the wheel is not being powered by the motor during launch and that the extra energy needed for the ball comes from the flywheel.   Also, friction and ball compression energy losses are assumed zero to simplify this analysis but can be significant in actual percentages derived.   I am focusing  on how increasing flywheel mass lowers the percentage range errors caused by ball mass variations.

E_wheel_initial/E_wheel_final = (1 + E_ball/E_wheel_final)

Lets expand E_ball/E_wheel_final

E_ball/E_wheel_final = (.7*m_ball*w_ball^2*r_ball^2)/(.5*I_wheel*w_wheel_final^2)

= 1.4*m_ball*w_ball^2*r_ball^2/(m_wheel*r_wheel^2*.84^2*(2*w_ball*r_ball/r_wheel)^2)

= .4954*m_ball/m_wheel

SInce  m_ball = 60 g and m_wheel = 180 g   m

_ball/m_wheel = 1/3

So  E_ball/E_wheel_final = .165    for a single 5″  wheel flywheel     .165/n for n flywheels.    So the ball energy is almost equal to the 1/6 final energy of the wheel

Range Tolerance analysis:

So how does R vary with m_ball from all this.   Well , we know the range is proportional to V^2 which is proportional to w_wheel_final^2 which is proportional to E_wheel_final.

From above E_wheel_final = E_wheel_initial/(1+ .4954*m_ball/m_wheel)

So due to proportionality of R and E_wheel_final we can say

R/R_0 = ((1+ .4954*m_ball_0/m_wheel)/(1+ .4954*m_ball/m_wheel))

where R_0 and m_ball_0 are the nominal values without errors.

We can use R range= R_0(1+ %e_r)   and m_ball = m_ball_0*(1 + %e_m_ball) to work with % changes.

Then with some manipulation we can get %e_r as a function of %e_m_ball

%e_r  = -%e_m_ball/(2.02*m_wheel/m_ball_0 +1 + %e_m_ball)

Now m_wheel = n*.180 kg   and m_ball= .06 kg  so we can write an approx.

%e_r = -%e_m_ball /( n*6.06 +1)     where n is the number of 5″ vex wheels.

Lets put in a few numbers:

Assume %e_m_ball = 10%  then the range error is

n = 1, %e_r =  -1.42%

n = 2, %e_r =  -.76%

n = 3, %e_r =  -.52%

n = 4, %e_r =  -.40%

n = 5, %e_r =  -.32%

So you see the benefits of having a higher  flywheel mass to ball mass ratio.   The use of  two 5″ wheels in a single wheel design can reduce a potential 10% range error from ball mass variations  to  1% ( less than a ball radius).   To keep the spin up time to a reasonable number of seconds requires about 2 393 motors per wheel so 2 wheels costs 4 motors.   So there is a motor tradeoff to get that  higher accuracy with heavier flywheels.

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Finding the moment of inertia of a Vex wheel using parallel axis theorem

May 17, 2015

vex 5 in wheel

The new Vex game “nothing but net” might involve rotating shooter wheels.  We know that if all the mass of a wheel was located on its rim then the moment of inertia about its rotating axis  (I_rim) would be

I_rim = r^2 * m   where m is the mass of the wheel and r = radius of wheel.   But we know that the wheels actually  have mass that is unevenly distributed along the radius so the moment of inertia I_wheel will be less than I_rim.

Easy experiment to determine I_wheel if we know its mass.

We can determine I_wheel experimentally using the parallel axis theorem and the dynamics of a pendulum.

Parallel axis theorem says that any object that is rotated about an axis parallel to and a distance , d, from an axis going through the centroid of the object will add an amount =  m*d^2 to the moment of inertia about its centroid.  I.e.

I_parallel = I_centroid + m*d^2 .

Suppose we now swing the mass, m,  about the parallel axis like a pendulum  using just the torque from gravity pulling on the mass.      It is easy to show that the period, T , of the pendulum is related to the distance , d, and the moment of inertial , I_parallel, by the following formula:

T = 2*pi*sqrt(I_parallel/(d*m*g))     .   g is gravitational constant and assumes swing angles smaller than say 10 degs from the lowest point on the pendulum path. 

If we measure the period of the pendulum we can rearrange the equation and find I_parallel

I_parallel = T^2*d*m*g/(2*pi)^2 

Once we have I_parallel, we can now use the parallel axis theorem to determine I_wheel.

I_wheel = I_parallel –  d^2*m  =  d^2*m * (T^2*g/d/(2*pi)^2 -1)

(This assumes  that the string has negligible mass relative to the mass of the wheel)

Vex 5 in Wheel experiment:

Given….r_wheel = 2.5 inches ,

wheel mass,   m = 180 gm (0.180  kg)

The pendulum is created by suspending the wheel with a thread 2.75 inches from its center so

d =. 07 m (approx. 2.75 inches)

The average period  T = .668 sec

I_wheel  = d^2*m*(T^2/d*9.8/(6.28)^2 -1)

     = d^2*m*( .248*T^2/d -1)

     = .07^2*.180*(.248*.668^2/.07 -1) = 0.00051 kg m^2

Equivalent radius with a rim only mass r_e

r_e = sqrt( I_wheel/m) = .0533 m ( 2.1 inches)  

This  means that the wheel behaves as if the mass if located at 84 %  of the radius of the wheel which one could almost guess by looking at it.