A friend of mine who also plays tennis was telling me the other day that his coach claims you can hit down and attempted to demonstrate doing so. My friend was not convinced that he was in fact hitting down.

Anecdotally I've heard you need to be 6'10" (2.08m) to get near hitting down, so I decided to do a bit of maths to see if I could work out where the contact point needed to be to hit your serve straight down into the service box. I decided to make some assumptions to make the maths a bit easier, and there are some things that to be honest would be too time consuming to include.

So, here we go.

The distance from the base line to the service line (BL to SL) is 18.3m. The net is 0.9m tall. The SL is 6.4m form the net.

A tennis ball is an average of 6.7cm in diameter. For the ball to miss the net by 1cm, the centre of the ball has to be 4.35cm above the net. That means the height from the ground to the centre of the ball is 0.945m. Using some simple trigonometry, the tangent of the angle at the net is 6.4/0.944 (6.77). If the distance from the BL to SL is 18.3m then by using Tan we can calculate the height of the contact point above the BL (18.3/6.77)=2.7m.

So, a ball hit down from 2.7m and following a straight line would miss the net and hit the opposite service line. But there's a snag. Well a few snags actually. First, there's gravity, then there's drag, and of course there's the speed at which the ball is struck. There's also spin that will affect how the ball moves through the air.

Let's deal with gravity first. Gravity will cause the ball to drop as it travels through the air. Gravity causes an object to fall at 9.8m/s

^{2}. So, from 2.7m it would take approx. 0.62secs to fall to the height of the net. But it's not simply being stopped, so that doesn't really help!

The distance from the contact point (2.7m above B/L) to the net along a straight line is 12.02m. If you hit a serve at 160km/h (100mph), it would travel that distance in 0.27secs. So, how far does the ball fall under gravity in 0.27secs? Well there's a formula for that:

d=gt

^{2}/2

That works out at 0.36m. That suggests that a ball hit at 160km/h in a straight line towards the opposite S/L will not clear the net!

At its simplest, moving the contact point up by 0.36m would mean the ball would clear the net, and because it continues to fall, it will land inside the service box. So now we have a contact point at 3.06m (10'3").

Drag will slow the ball down as it flies through the air. From what I've read, and again without lots of verified data, it seems as though the accepted impact is that by the time the ball reaches the S/L it has probably slowed down by 60%. As the ball slows down, the impact of gravity will be more significant. In other words it will drop further for each unit of horizontal distance it travels as it decelerates.

If you stayed with me so far (and remember we're trying to keep this as simple as we can), let's assume that the ball slows down at an even rate as it travels. Using those simple assumptions, deceleration will allow the ball to drop 0.045m. Again, keeping things simple, that moves our contact point to 3.11m (10'5").

I'm 6'3" and wth my arm fully extended the sweet spot of my racquet is not high enough to reach that point and create the correct angle. With my normal service action I usually leave the ground, so it's probable that I get my racquet high enough, but only just, to do this, and I need to hit the ball at 100mph to do so! What happens if you slow the serve down to say 80mph? Well I could do the maths, but I'm not going to. The obvious answer is the contact point will need to move higher to compensate. If everything is directly proportional then that could be as much as another 2' of height (60cm). In other words, you'd need to be 8' tall!

The simple conclusion to all of this is that when your coach tells you that you need to strike up through the ball, then trust him/her. They ay not do the maths, but the understand tennis!!

## No comments:

Post a Comment