# Why do ten-pin bowlers prefer heavier bowling balls?

Why do ten-pin bowlers prefer heavier bowling balls?

From a physics perspective, it seems like the best way to be sure to knock all of the pins down would be to transmit the most energy to the ball/pins system. From the calculation for kinetic energy, $$T \propto m v^2$$ we can see that increasing the velocity would be much more effective than increasing the mass. Since you can throw lighter balls faster, it would make more sense to throw a light ball fast, rather than throw a massive ball more slowly.

I'm not the best bowler, so I'm not always very accurate with my throws. This strategy works better for me and I get more strikes by throwing a light ball fast. What differences are there at the professional level that might cause all these factors to change, making the heavier ball the better choice?

• Interesting thought... So a pitcher might throw baseballs at bowling pins... Oct 30, 2016 at 17:20
• I suspect it's due to precision. If you have a very light ball then, as your arm is swinging, the slightest error in alignment - the slightest twist of the wrist - will be transferred to the ball. Whereas the inertia of heavier balls are less prone to such erroneous perturbations. Oct 30, 2016 at 17:20
• A good answer to this question would consider whether it's more important to transfer energy to the bowling pins, or momentum.
– rob
Oct 30, 2016 at 17:59
• The pins are not as light, as one might think. I've seen a bowling ball "severely" change direction at collision. I will suspect that the mass is a key factor in being sure that the obtained path, curve, rotation and speed of the ball remain unaltered while colliding with all the 10 pins and aren't changed and reduced at the first pin collision. Oct 30, 2016 at 18:12
• I wonder if this is a little bit of "misdirection" - perhaps the actual reason bowlers like heavier balls (assuming they do) is not related so much to the ball-pin dynamics, as it is to ball-lane and/or ball-air dynamics. I know when I try to bowl, the ball never even gets near the pins because it doesn't roll in the direction I'm hoping for.
– uhoh
Oct 30, 2016 at 18:48

The main things to consider when bowling are:

• you want the ball to go in the direction you intend to
• after a collision:
• the pin must fall down (receive enough momentum/velocity)
• the trajectory of the ball should remain more or less the same
• there must be enough kinetic energy left to hit the other pins to the ground

As stated in one of the comments, a heavier ball is superiour on the first point, because a slight swing of the wrist has a larger influence on the lighter ball. Of course, the ball can't be that heavy that you struggle to swing your arm, because then, the accuracy gets worse instead of better.

On the next points, the heavier ball is always better. To put it clear: when a truck hits an apple, the apple will notice and significantly change velocity, while the truck is hardly influenced and has enough kinetic energy left to knock some other apples away.

In more accurate terms:

You give the ball some kinetic energy in order to roll towards the pins. A part of that kinetic energy is transferred to the pin it collides with in order to tip it over. The next graph depicts the percentage of the original kinetic energy that is not transferred to the pin and thus is still available to knock over other pins (assuming an elastic collision) (for the pin, a mass of 10 was assumed):

This maximizes for a ball with mass 0. A ball of weight 0, however has the clear disadvantage of not being able to knock anything down. After becoming zero when both ball and pin have the same weight (the pin gets the velocity of the ball and vice versa), it rises again with increasing ball-weight.

To prove this:

let's say, independent on the ball's weight, the kinetic energy it starts with is always $$T$$. Before the collision, the ball has a velocity $$u_1$$ and the pin velocity 0. The velocities after are $$v_1$$ and $$v_2$$.

\begin{aligned}T &= \frac{m_b u_1^2}{2} \\ \Rightarrow u_1 &= \sqrt{\frac{2T}{m_b}} \end{aligned}

For the velocity of the pin after the collision, this gives:

\begin{aligned}v_2 &= \frac{2 m_b }{m_b+m_p}u_1 \\ &= \frac{2 m_b }{m_b+m_p}\sqrt{\frac{2T}{m_b}} \\ & = \frac{2 \sqrt{m_b} }{m_b+m_p}\sqrt{2T} \end{aligned}

If the mass of the ball is small with respect to the mass of the pin, the velocity the pin has after the collision is far too low to fall over. This can be seen in the next graph as well. It depicts the velocity of the pin after the collision for a fixed T to start with.

Combine this with the graph above and it is clear that a heavier ball is better.

(ps: A mass lower than the mass of the pin is a bad idea for a different reason too: supposing an elastic head on collision, the ball will bounce backwards and won't hit any other pin)

Edit

For a fixed amount of kinetic energy to start with, it's not just the heavier the better: quite a lot of pins will not be knocked down by the ball itself, but by other pins. To maximize this, the velocity of the pins needs to be large. As the ball gets heavier, the starting-velocity will decrease and the velocity of the pin after the collision as well (see second graph). For that pin itself it won't matter that much: it will fall. With lower velocity, the probability of that pin knocking down a next pin, however, will be lower.

First, I will suppose that the bowler can supply a constant kinetic energy $$k$$ to the ball regardless of its mass, and will search for the optimal mass $$m$$ for the ball that transfers to a pin with a known mass $$M$$ the largest possible speed for the energy $$k$$.

For the sake of simplicity, lets consider a head on perfectly elastic collision (we'll get the same result with a non elastic collision but the calculus would be messier).

Conservation of linear momentum yields : $$m.v_1=m.u_1+M.u_2$$

$$v_1$$ is the velocity of the ball before collision, $$u_1$$ velocity of the ball after collision, $$u_2$$ velocity of the pin after collision ( supposing it was at rest before ).

Now perfectly elastic gives :$$v_1=u_2-u_1\implies u_1=u_2-v_1$$

Substitute $$u_1$$ in the first equation and solve for $$u_2$$:$$u_2=\dfrac{2m}{m+M}v_1$$

Now using the bit of constant initial kinetic energy:$$k=\dfrac{1}{2}mv_1^2 \implies v_1=\sqrt{\dfrac{2k}{m}}$$

Therefore,$$u_2=\dfrac{2m}{m+M}v_1=\dfrac{2m}{m+M}.\sqrt{\dfrac{2k}{m}}$$

which simplifies into :$$u_2=c.\dfrac{\sqrt{m}}{m+M}$$

where $$c$$ is a constant. The curve of $$u_2=f(m)$$ looks like this :

It reaches its maximum at $$m=M$$, so, if you where shooting at a constant energy, the optimal mass is that of the pin.

Well now if we supposed that the player can apply a fixed momentum at the beginning ( instead of fixed energy ), similar calculation will lead us to :$$u_2=\dfrac{c}{m+M}$$which is clearly maximum for $$m$$ minimum, though smaller masses suffer more energy loss on the way, so you'd have to find an optimum somewhere in the middle.

• choosing the mass of the ball between zero and the mass of the pin (as I understand you propose) is a very bad idea: if the ball and the pin make a head on elastic collision, the ball will bounce backwards and hit no other pins. That would make it very hard to throw a strike Oct 30, 2016 at 22:39
• @Dries i didn't say this, the mass of the pin is usually 1.6 kg whereas that of the ball can be as large as 7.26 kg. I said that a lighter ball can give more momentum to the pin, though i think you're right on the point of keeping energy in the ball for other pins.
– Tofi
Oct 31, 2016 at 4:56
• Great analysis, but I'm having a little difficulty with your first assumption, that the energy transmitted is constant. Can you explain why you made that assumption? Oct 31, 2016 at 16:40
• @psitae the first assumption was that the energy given to the ball when delivering it is the same every time; this happens if the player applies on different balls the same force over the same distance. The second assumption was that the ball has same momentum every time; this happens when the player applies on different balls the same force over the same time. So it depends on the way you shoot the balls.
– Tofi
Oct 31, 2016 at 19:39