Energy ratio method for quantifying effect of ball mass variations on range error of a flywheel ball shooter

The new Vex game “Nothing but Net” requires a ball to be shot across the field into a goal.   The ball has a mass specification that allows up to 10% variation.    I derived equations that related the range error to mass error for a single flywheel shooter here and wanted to generalize the analysis to any flywheel system that uses the energy of the flywheel to impart energy to the ball.    It turns out that the error in range due to mass is a simple function of the ratio of flywheel energy to the energy of the ball.

i.e. %range error = %mass of ball error /(1+ ratio)

where ratio = flywheel energy/ ball energy.

Following is a derivation of this formula.  To simplify the work I will define a term called equivalent mass, Meq,  for the flywheel which is a mass that would give the same kinetic energy of the flywheel if it were traveling linearly at the ball speed, V.

I. e.   we can write the flywheel energy , Ew = 1/2*Meq*V^2

We know that Ew =1/2* I_wheel*w_wheel^2 so

So  Meq = I_wheel*w_wheel^2/V^2 

For a single wheel shooter, w_wheel = 2*V/r_wheel so

Meq_1 = 4*n*I_wheel/r_wheel^2

where n = number of wheels.

For a two wheel shooter, w_wheel = V/r_wheel  so

Meq_2 = n*I_wheel/r_wheel^2

If we write I_wheel in terms of its radius and mass then

I_wheel = c*m_wheel*r_wheel^2.     Substituting in the equivalent mass equations gives

Meq_1 = 4*n*c*m_wheel

Meq_2 = n*c*m_wheel

.Ball Energy, Eb

Eb_1 = 1/2 *m_b*V^2*(1.4)   (1.4 factor is for added spin energy)

Eb_2 = 1/2*m_b*V^2

Conservation of Energy

The initial energy of the flywheel is equal to the final energy of the wheel plus the ball energy.

Ew_initial = Ew_final + Eb

Ew_initial = Ew_final*(1+ Eb/Ew_final)  = Ew_final*(1 + 1/ratio)


ratio = Ew_final/Eb

Lets look at Eb/Ew_final

single wheel shooter

Eb_1/Ew_final_1 = 1/2*m_b*V^2*1.4/(1/2*Meq_1*V^2)  = 1.4*m_b/Meq_1


ratio_1 = 5/7*Meq_1/m_b

two wheel shooter

Eb_2/Ew_final_2 =  1/2*m_b*V^2/(1/2*Meq_2*V^2)= m_b/Meq_2 so

ratio_2 = Meq_2/m_b

So it looks like  ratio is only a function of m_b and not a function of V.

Rewriting the energy equation to isolate Ew_final

we get

Ew_final= Ew_initial/(1 + 1/ratio)

Since Ew_final is proportional to V^2 which is proportional to range R

we can say that

R/R_0 = (1+1/ratio_0)/(1+ 1/ratio)

and introducing % changes   R=R0 *( 1 + %e _r), m_b = m_b_0*(1 + %e_m_b)

we can after some manipulation show that

%e_r = %e_m_b *( 1/(1+ ratio) = %e_m_b*factor

where the  mass % error reduction factor=1/(1+ ratio)   and  ratio . is defined above for each type of shooter.

factor vs ratio is simple to compute

ratio, factor

.5 , 2/3

1, 1/2

2, 1/3

3, 1/4

4, 1/5

5, 1/6

9, 1/10

Single wheel shooter

ratio_1=  Meq_1/1.4/m_b_0= 4*n*c*m_wheel/m_b_0/1.4

Two wheel shooter

ratio_2 = Meq_2/m_b_0= n*c*m_wheel/m_b_0      

Range performance

Lets put some numbers for the 5″ vex flywheel.

m_wheel = .18 kg, m_b_0 = .05 kg, c = .84^2 = .71

Single wheel shooter factor:

ratio_1 = 4*n*1.79/1.4

ratio_1 = 4*1*1.79/1.4 = 5.1   => factor = .163  (assumes 1 wheel for flywheel)

ratio_1 = 4*2*1.79/1.4 = 10.2   => factor = .089   (assumes 2 wheels for flywheel)

Two wheel shooter factor:

ratio_2 = n*1.79

ratio_2 = 2*1.79 = 3.6    => factor = .217 (assumes 1 wheel on each side)

ratio_2  = 4*1.79 = 7.2   => factor = .12  (assumes 2 wheels on each side)

ratio_2  = 6*1.79 = 10.74  => factor = .085  (assumes 3 wheels on each side)

Observation…. although the two wheel shooter flywheel has 1/4 the energy due to its slower speed, the ball energy is also less because it requires no spin energy. The energy ratio is only different by a factor of 2.85 rather than the full 4.  Still, the single wheel shooter has about 1.3% less range error than the two wheel shooter with the  n=2 configuration.   This is not that big a deal since we are talking 1.4 inches error (compared to a 4 inch ball diameter).


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