What I'm saying is that if the ball is thrown at EXACTLY the same speed over the EXACT spot from the EXACT same spot, with no variation at all. Then the ball's path and deflection would be EXACTLY the same.
Yes, granted if you repeat the shot identically in this ideal scenario it will strike every time. The question was, wether throwing exactly the same shot
faster is any different. Obviously with the 6lb ball the answer is "Yes, you can achieve a strike with the faster ball from a line that might otherwise deflect significantly at a slower speed"
To the other guys; F=ma is completely irrelevant, this equation will tell you how much FORCE your hand (cannon) imparted on the balls MASS to achieve a given ACCELERATION onto the lane. OR you could use the FORCE of friction to tell you how much the ball will decelerate, if you knew the value of F(friction) on a lane.
Momentum is the key, and potential energy. P=mv. This simple equation flatly states that the faster you deliver the ball the MORE ENERGY it will have when it reaches the pins (yes I know we all know that already). Given ten equally weighted pins set identically rack after rack, the requirement of P is constant. We introduce the next variable m by choosing the mass of the ball. So given P and m, we can derive the required speed (v) that the ball of weight (m) must be travelling to impart the required energy (P) onto the pins.
Now that we know we have MORE ENERGY with a FASTER ball, other variables need to be considered such as actual position of the shot.
For example, if you are in the exact spot required that all of the pins cannon into each other for a strike, the amount of energy is irrelevant ONCE YOU ACHIEVE the required amount and above. Any faster will achieve no different result.
If your placement requires a messenger to trip a corner, the energy (P) imparted upon the pins will have a range, because as we all know if you hit the deck too hard you can push the pins off the back before they have a chance to move both ways across the deck. THEREFORE your speed (v) must be between x and y for the messenger to work. Too slow (and therefore lower P) and the messenger may not reach its target, too fast and the messenger might disappear before that trip.
Alternatively if the line requires the ball to carry through the 8pin (right hand) then P must be sufficiently high that deflection as mentioned previously does not exceed the angle to the 8pin. Assuming an inelastic collision you could calculate the vector of reflection (given we know P, and can calculate the force required to move a pin) In this circumstance, faster is better up until you achieve a high enough speed (v) to counter the differential. Once you reach this point, the equations revert to the scenarios above (messengers and carry)
So there's a lot more variables happening, but I would say in conclusion that it's the balance between force and placement that achieves optimum result. Neither of these variables can dominate on its own.
In practice of course, there is actually a bit of a bell curve; the quicker bowlers (not the fastest) tend to score better in my experience.
For those that are interested, there was actually a bowling question in the NSW Physics HSC exam when I was doing it.
