## 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.

## Waypoint steering geometry for a mobile robot

February 5, 2013

Fig 1 shows the geometry that I will use in the waypoint steering algorithms that will be discussed in future posts.  The field geometry is North (N) along the y-axis and East (E) along the x-axis.   The heading angle , psi is defined positive clockwise from North as a compass rose.

Waypoint:   The waypoint is defined with both location (X_p, Y_p) and direction psi_p.   The slope of the center track line m is  1/tan(psi_p).

Cross track and Along track Ranges:

The two parameters of interest are the along track range R_y and cross track range , R_x.   The steering is determined by R_x = R*sin(delta_psi) where delta_psi is the angle between a line drawn from the robot to the waypoint and the actual track line.

Derivation of R_x  and  R_y without sin and cos functions:

We know that the track can be expressed as a line that goes through (X_p,Y_p) and has a slope m.   So we can write the track line as  y= mx + y_0 form but we need y_0.    Plugging in for point p gives  y_0  = Y_p – m*X_p .

Hence the line eq  is   0 =  -y +m*x + Y_p – m*X_p.

It can be shown that the distance R_x  between a point  (x_r, y_r)  and a line  a*x + b*y  + c  = 0  is

|R_x|= | (a*x_r+b*y_r +c)|/sqrt(a^2 + b^2)    (see  wikipedia)

where a = m ,  b= -1 ,  c= Y_p – m*X_p ,    m  = 1/tan(psi_p)

This form can lead to division by zero when psi_p = 0 but can be avoided by using the tan(90 – psi_p)  = 1/ m  .  We  divide the starting equation

0 =  -y +m*x + Y_p – m*X_p

by m   to change its form and use it whenever m >1 or psi_p > 45 degrees.

0 = -y/m + x + Y_p/m – X_p.

Now we have new definitions of a,b,c

a = 1, b = -1 /m, c = Y_p/m – X_p  , 1/m = tan(90 – psi_p).

R_y derivation

We can use a similar method to find R_y.

|R_y| =  |(a’*x_r+b’*y_r +c’)|/sqrt(a’^2 + b’^2) )

where we use the perpendicular line to the track passing  through (X_r,Y_r) with slope  m’ =-1/m and the point as (X_p,Y_p).    This leads to

a’ = -1  , b’ = m’ , c’ = Y_r  – m’*X_r  , m’ = -1/m = -tan(psi_p)  .  We use this when psi_p  < 45 degrees and the modified form

a’ = m , b’ = 1 , c’ = -Y_r*m – X_r    for psi_p > 45 degrees.

Making R_x and R_y signed values

By convention, I want displacement to the right as positive R_x.  (X_r,Y_r) will be to the right of the track if  0< delta_psi  < 180 otherwise we change the sign of R_x.

Similarly, we want the along track distance, R_y,  to be positive if we are heading in the direction of the waypoint along the track and we have not reached the waypoint. This occurs when  0<90 – delta_psi < 180 otherwise R_y is negative.