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

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.

Note: Water Jet Physics.. Equivalent Nozzle Diameters (Dual vs Single)

November 15, 2013

Question:  What is the diameter of a single nozzle that has equivalent thrust to two smaller diameter nozzles being fed from the same input source (say a skyboard 4 in fire hose)? 

I reviewed the physics of water jets in this post for an inventor friend Bob.   He posed the above question to me since he was converting a dual jet concept to a single jet and wanted the same thrust.

I gave the following answer:   Continuity requires that the total mass flow rate exiting  both dual jets must equal the input mass flow rate before the split.   Momentum equations (spoken about in the referenced post) state that the total thrust T for a stationary jet nozzle is

T = Md*Vexit   

where Md is the water mass flow rate and Vexit is the water exit velocity.      Vexit must be the same for both the single and dual jets.   Using two jets just splits the mass flow rate into two streams each with mass flow rates of Md/2 but Vexit must remain the same for both cases to generate the identical thrust.    I.e.  the dual nozzle case looks like this:

T = (Md/2)*Vexit + (Md/2)*Vexit   (dual jet case)

= (Md/2 + Md/2)*Vexit  

= Md*Vexit   (equivalent  single jet case)

Intuitively we can see that if the exit velocity, Vexit, is the same for both cases then the total areas ejecting water must be the same to get the same mass flow rate.  The area of the single nozzle must be twice the area of one of the dual nozzles.   And since area is proportional to the square of the diameter then the diameter of the single nozzle must be the √2 * the diameter of the smaller nozzle.

How is the diameter calculated given  required thrust T and mass flow rate Md?

First calculate Vexit

Vexit = T/Md

Substituting Vexit into the mass flow rate  equation Md = rho*A*Vexit  where rho is the density of water  and solving for A gives

A = Md/(rho*Vexit) = Md^2/(rho*T)

And finally knowing A we can solve for the diameter:

Diameter =2* √(A/π)

Water jet Flyboard thrust physics (rev 4.26.13)

March 27, 2013

zapata jet pack

I met an inventor named Bob at the Long Beach FRC robotic regional competition who wanted my opinions on how to best stabilize single nozzle version of the  Zapata racing Flyboard which was inspired by the JetLev.  Another newer version of this is called the Jetovator.  All three are summarized in this Gismag article.

I wanted to refresh myself on the physics of the thrust that drives the jet pack.

We know that  F = d(mv)/dt   which under steady speed  conditions and equal input and output pressures leads to the thrust  (T) equation (ref NASA thrust site)

T = F = m_dot*(v_e – v_i)

where m_dot= time rate of change of  water mass flowing through a hose  and

v_e = water exit speed , v_i = water input speed along the same direction.

The mass continuity equations yields

m_dot = A_hose*rho*(v_hose)   = A_nozzle*rho*v_e  

where rho is the density of water and A_hose is the area of the hose, v_hose is the water flow rate in the hose and A_nozzle is the area of the thruster nozzle.

Since v_i = 0 for a stationary jet ski pumping water vertically to the flyboard then thrust equation for T in terms of v_e gives

T = A_nozzle*rho*V_e^2

If the hose and nozzle areas are different the relationship between v_hose and v_e is

v_e = v_hose*A_hose/A_nozzle. 

Substituting into the T equation for V_e give T in terms of v_hose.

T = A_hose*rho*v_hose*v_e = (A_hose*v_hose)^2*rho/A_nozzle

or in terms of hose flow rate  q   = A_hose*v_hose

T = q^2 *rho/A_nozzle

Other data we will need:

A_hose = 1/12 sqft (approx 4 in hose diameter),

g = 32.2 ft/s/s,

rho = 1.94 slug/cft

Under steady state flight the total thrust must equal to the weight.    It is assumed that the water weight is supported by the thrust since it puts a downward force on the hose directly beneath the vertical water column.     The thrust must also support the dry weight of the hose extending above the water.  (Note:  the hose has viscosity drag due to the high-speed water flowing through it.  The reaction force adds lift to the side of the hose. but is cancelled by the opposing downward force on the bottom of the hose.)

T = W_platform + W_hose_dry + W_water 


W_hose _dry= 85 lb*H/100ft    (4 in standard fire hose)

W_water= g*rho*A_hose*H

and H is the length of the hose above the water.

Substituting the weights into the T equation above we can rearrange terms and solve for q

q = sqrt((W_platform + W_hose_dry + W_water)/ (rho/A_nozzle ))

Comparison of Bob’s Test numbers with Theoretical calculation:

Bob built a test platform which weighed 250 lbs with him aboard.  At a height of 30 feet the exit flow rates were around 1000 gpm and the jet ski ran at 7100 rpm.   At 5 ft , the ski jet rpm lowered to 4500 rpm.  

The  rpm ratio he saw was  7200/4500 = 1. 6 between heights.   He said he had two nozzles with a diameter of 1.7 inches (A_nozzle =2* .0158 sqft= .0316 sqft) . 

The total length of these hoses are typically 50 ft when you account for the portion of the hose in the water.

So lets get started with the calculation based upon my equations:

The dry weight of the hose at 30ft  and the water weight are:

W_hose_30ft = 85*30/100 = 26 lbs    ,   W_water_30ft = 156 lbs

W_hose_5 ft  = 4.25 lbs                          , W_water_5ft = 26 lbs

Lets calculate the required  flow rate  q at 30 feet  and 5 ft  and see how it compares with Bobs.

q_30 =   2.65 cfs ( 1189 gpm )   , q_5 = 2.13 cfs (958 gpm)

The flow ratio  = (q_30/q_5) =  2.65/2.13 = 1.24

The calculated flow rate of 1189  gpm compares within  19% of the observed 1000 gpm .I would expect a little better agreement… I am not sure to what accuracy Bob estimated his observation or how it was measured.

RPM ratio :

Normally, pump affinity laws state that flow rates are directly proportional to rpm’s and we could stop here but as you see the flow ratio is 1.24 compared to the observed rpm ratio of 1.6.   This is because there is static pump pressure and friction losses as well as dynamic pump pressure.   So it is more accurate to compare the total pump pressure at each height and use a pump affinity rule that says the rpm is proportional to the sqrt of total pump pressure.

To calculate the rpm ratio of the pump we need the pump head or pump pressure p_p.  Once we have that on the assumption of a centrifugal pump we know that the rpm’s are proportional to the sqrt(p_p) assuming the same pump efficiency.   Using Bernoulli’s total pressure relationship we know that the total pressure at the pump must be higher than the total pressure at the nozzle  by the added pressure of the water weight (p_h) and the friction back pressure (p_fl).

p_p = p_f l + p_T + p_h

 To compute  p_fl   I  assume that the total hose length is 50 feet and doesn’t change with how much hose is above water.

I found an internet equation  (ref)  that gives the hose friction pressure FL in terms of lbs per sqin.

FL _lbs_per_in2= C*(q_hose_gpm/100)^2*L_ft/100   where

C = Friction loss Coefficient  = .2   for a 4 in fire hose.

To keep units the same we can write p_fl  in terms of q (cft/s)

p_fl =144*FL =  C’*q^2*50ft    with C ‘ =  C*(448.8)^2*144/(10^6)=  5.8

 p_T  = T/A_nozzle  is the pressure on the nozzle area due to the thrust and is the total pressure at that point

p_h  = W_water/A_hose  is the pressure due to the water height

 Pressure ……………………   H_ft = 30 ft……………………..H_ft = 5 ft










 p_p (sum)



 Recall that assuming the same pump efficiency, the rpm’s are proportional to the sqrt of the total pump pressure.  So

rpm_ratio = sqrt(p_p_30/p_p_5) = sqrt(26548/11037) = sqrt(2.4) = 1.6

Amazing!  This is matches the observed rpm ratio of 1.6.