The new vex game , *Nothing but Net*, could utilize a design similar to a two wheel tennis ball launcher.

**Question: how many motors are required on the launcher?**

The after a launch, the energy lost from the spinning wheels is transformed into ball kinetic energy and heat due to friction and ball compression.

After each shot the wheels are brought back to initial spin speed by the power of the motors. The maximum time allowed for respinning is the cycle time of the firing sequence. Lets take a look at the Vex game derived requirements:

Ball mass, **m = 60 grams**

Ball launch Speed, **v = 6 m/s**

Ball kinetic energy:**K = 1/2*mass*v^2 = .5*.06*6^2 = 1.08 joules**

Energy loss due to compression : **E_c**

Energy loss due to friction : **E_f**

Time between shots: **1 sec**

Average power required **p_avg = ( K + E_c + E_f)/ t**

**Force on ball during acceleration:**

F = d(m*v)/dt or the change in momentum of the ball over the time of acceleration., dt.

dt can be approximated as the contact distance / tangential speed of the wheel , v.

The **contact distance is about 3 cm** so

**dt= .03/6= .005 s**

**d(m*v) = .06*6 = .36 kg*m/s**

hence **F = .36/.005 = 72 newtons**

Normal force on ball **F_n = F/u_friction . ** The normal (compression) force on the ball is then

F_n = 72 newtons assuming a u_friction = 1 which is possible with a sticky wheel.

The assumed compression distance is about 1 in or 2.54 cm. (To be verified later)

Hence **Ec = F_n*d/2 = 72*.0254/2 = .91 joules. **

With good design, the **friction loss in the drive train** can be small (maybe **.1 joules**) so lets assume that E_c + E_f are about equal to the K= 1.06 j so

**p_avg = 2*K/t = 1.06*2= 2.12 watts or 1.06 watts per motor.**

We know the **vex 393 motors** have a max power = max_speed*max_torque/4 or about **4.5 watts** but they will overheat if run continuously at this power. The PTC fuses will stop the motors if they run continuously with currents equivalent to more than 25% maximum torque (speeds less than 75% max speed). At this operating point, the motors only deliver 3/4 max power or** 3.4** watts.

There are also friction losses from the teeth of the spur gears. I usually assume about **5% per 5:1 ratio**. A shooter wheel with a 25:1 gearing would lose 10% torque or energy at a given speed.

So the net power to the shooter wheel will be .9*3.4 = **3.0 watts which is more than the 1 watt that we require. **

**Extra friction? **

If the gear train has pressure on the axles from bearing blocks and possibly the collars are too tight so the wheels slow down quickly when coasting with motors disconnected, then over heating can easily occur. e.g. if friction uses up just 15% of the available torque, the motors will have to provide about 1 watt extra. which cuts our margin considerably.

**Faster shot rate?**

We assumed 1 shot per second…what happens with 2 shots per second…. Well, the power requirements almost double since we are using twice as much energy per unit time. We would likely have to add extra motors.