Sounds too good to be true? Think I'm joking? Nope, I'm perfectly serious, but with one small proviso (there's always a catch!) - you need to be carrying some extra weight. Around 11 pounds (5kg) in fact.
How To Increase Your Footwork Speed - The Facts and Figures
Here's the skinny(!). Those of you who read my personal table tennis blog would know that at the time of writing (May 2008) I'm trying to drop weight to get ready for the Australian Open. I got to wondering how much faster I would get for every kilogram of weight I lost. Would it really be a significant improvement, or would it be hardly noticeable?After finding the physics formulas, I played around with the math and came up with some interesting conclusions. I'll include the figures here for reference, but those of you who detest numbers can jump to the conclusion at the bottom of the page.
Getting Started
First of all, I needed to make a few assumptions to give me a starting point. So I decided to crunch the figures for a 165 pound (75kg) male, who is going to lose 1kg (2.204 pounds) of bodyfat, with no muscle loss. I then checked out some video footage of table tennis matches, and found that in a typical looping/counterhitting rally there is around half a second between each stroke of the rally, so each player has about 1 second between each of his strokes. I then tested myself moving sideways from a stationary ready position, and found that I could move 1.60 meters in 1 second. Armed with this knowledge, I was able to do some rough and ready calculations.
The two physics formulas I need are the formula for working out the acceleration of an object, given how far it moved from a standing start in a certain time, and how much force is needed to produce that acceleration in an object of a certain mass.
Assuming a person generates a certain amount of force to move a distance in 1 second, if that person loses 1 kg of bodyfat without losing any muscle, he should be able to generate the same amount of force. But since he is lighter, his acceleration should increase, allowing him to move further in 1 second.
The relevant formulas are as follows:
d = ut + ½at2 ; and
F = ma; where
d = the distance moved by the person (in this case 1.60 meters);
u = initial velocity of the person (in this case zero, since he is stationary at the start);
t = the amount of time the person was moving (1 second);
m = mass of the person (75kg);
a = acceleration of the person (which we want to find out); and
F = force generated by the person (which we want to find out).
First, let's work out the acceleration of the person.
d = ut + ½at2
1.6 = 0 + ½a
a = 3.2 ms-2
Now, let's find out the force the person generated in order to move with that acceleration.
F = ma
F = 75 x 3.2
F = 240 Newtons (N)
Now that I knew the force generated, I assumed the player lost 1 kg of bodyfat, with no muscle loss. Therefore he should be able to generate the same amount of force - 240N. But since his mass is less, his acceleration should increase.
F = ma, thus a = F / m
a = 240 / 74
a = 3.243ms-2
So we can see the player has increased his acceleration a little by dropping weight. This increased acceleration will allow the player to move further in that 1 second available, and also to cover the 1.6 meters in less time. Let's have a look at the results for both cases.
Distance Covered In 1 Second
d = ut + ½at2
d = 0 + ½ x 3.243
d = 1.6216 meters
Since the player could originally move 1.6 meters in 1 second, he has increased his court coverage by 2.16cm. But remember that this is only in one direction, so in fact the player has increased his full court coverage by 4.32cm overall (2.16cm to the left, and 2.16cm to the right). Given that his original court coverage was 3.2 meters ( 1.60 meters in each direction), this means that his percentage improvement is actually
% improvement = (4.32cm / (320 cm) ) * 100
% improvement = 1.35%
Not a bad improvement for a 1 kg drop in weight!
But what if the player dropped 11 pounds (5kg), with no muscle loss?
a = F / m
a = 240N / 70kg
a = 3.429ms-2
Distance Covered in 1 Second
d = ut + ½at2
d = 0 + ½ x 3.429
d = 1.7143 meters
Now our player has increased his court coverage in one direction by 11.43cm, and 22.86cm overall - or ¾ of a foot! This means his percentage improvement is now
% improvement = (22.86cm / 320cm) x 100
% improvement = 7.14%
A 7.14% improvement in court coverage - now we are talking!
Let's imagine that the player has a ball that is 1.6m away from him that he has to reach. Currently it would take him 1 second to reach the ball. How much faster could he get there if he lost 1kg of bodyfat? Let's find out.
d = ut + ½at2
1.6 = 0 + ½ x (3.243)t2
(1.6 x 2) / 3.243 = t2
0.987 = t2
t = 0.993 seconds
So our player can get to the ball 7 milliseconds faster. Now what if the player loses 5kg? We have already determined above that his acceleration would improve to 3.429ms-2, so he should get to the ball even faster.
d = ut + ½at2
1.6 = 0 + ½ x 3.429t2
(1.6 x 2) / 3.429 = t2
0.933 = t2
t = 0.966 seconds
So the player can get to the ball 34 milliseconds faster. Considering that the average reaction speed is considered to be around 300 milliseconds, having an extra 34 milliseconds to prepare yourself is not a bad thing!
