There are two main pieces to the puzzle:
- Determine the effect on take off velocity.
- Determine the impact of that change in velocity to vertical jump (the easy part).
Introduction to the Easy Part
The classic equations of kinetics will help us with 2, in particular the equation that describes the relationship between displacement and time under constant acceleration. Earth's gravity exerts an amount of force on all bodies in proximity to it which results in the same downward acceleration (in the absence of other forces). The force isn't constant for all objects, but the acceleration is. If you start from a constant acceleration of \(g = -9.81 m/s^2\) and you know the initial vertical velocity of your take off, you can arrive at the classic velocity and displacement formulas:$$v(t) = v_0 + g t$$
$$d(t) = d_0 + v_0 t + \frac{1}{2} g t^2,$$
where \(d_0\) is initial displacement and \(v_0\) is the initial (vertical) velocity. I was too lazy to look that up, so I just derived it. Note that I have put the minus on the gravity and not in the formulas. We can simplify this down to the variables we care about based on the fact that the velocity at peak (time \(t_p\)) is 0:
$$d(t_p) - d_0 = -\frac{v_0^2}{2 g}.$$
You'll notice that it doesn't matter how much you weigh for this part of the discussion. It matters how fast you are moving vertically at the moment your feet leave the ground. Now, in order to measure your vertical for the purposes of determining your initial velocity, you need to have a starting reach and a final reach. Your starting reach \(d_0\) will be taken on your tiptoes, because your foot extension is still imparting force to the ground. (I don't enough about this topic to say how critical this phase is, but I did a little self experimentation and my limited self-awareness suggests I'm using foot extension as a part of my jump.) You're going to need your peak reach on the tape as well. My displacement under this manner of measurement is 17" = 0.43m (but for the record my standard reach vertical is 20"). My initial velocity is about \(2.91m/s\). Does this help with determining my vertical if I lost 20lbs? Not really. But it's a cool number to know.
1. The Hard Part
The math we will do here is not difficult, but the justification is a little hazy. We need some kind of a model that explains the relationship between my weight and my initial velocity in a vertical jump. I'm going to model the process of jumping as having near constant acceleration. This is probably bogus, but I want a number. Let's put all the cards on the table:- Near constant vertical acceleration.
- The force production of the legs is the same for the heavy me and the light me.
These assumptions are the basic ingredients of a creamy fudge. Assumption 2 is probably okay for small enough changes in weight, but note that large changes in weight would invalidate it. You can convince yourself of the limited applicability of assumption 2 by attempting to do shot put with a softball and a 12 lb shot. You'll be underwhelmed by the distance the softball goes if you haven't already intuited it.
Measuring the distance of the vertical acceleration is a bit of a problem. Consider that your feet are not accelerating for the first part of the movement, but your head is. We're going to need to know the distance that the center of gravity is moved through the movement. There is some research that has gone into the question of the center of gravity of a human body in different positions. Here's a quick and easy to follow presentation of the topic from Ozkaya and Nordin of Simon Fraser University. What's missing is the magic numbers I want.
Physiopedia says the CoG of a person in the anatomical position "lies approximately anterior to the second sacral vertebra." Another resource (from Peter Vint, hosted on Arizona State University website) indicates that this is located at about 53-56% of standing height in females and 54-57% of standing height in males. I'm going to need to tweak with this to get my take off CoG. I also need my bottom CoG height. I'm going to be a little more vague about coming up with that number. I'm going to eyeball it with a tape measure and reference some diagrams I found online.
55.5% of my height would be about 39". Add to that the height of going on tiptoes (3") to get 42" height of CoG at take off. I estimate my initial CoG at the bottom of my crouch as 27". So, I have 15" = 0.38m within which to accelerate. We need to bring some equations together here:
$$v_f = a t$$
$$d = \frac{1}{2} a t^2,$$
where \(d\) is the displacement of my CoG from crouch to take off, \(a\) is my net acceleration and \(v_f\) is my velocity at the end of the jumping phase (take off). This is the same number as my initial velocity (earlier called \(v_0\)). Solving for \(a\) yields
$$a = \frac{v_f^2}{2 d} = 11.1 m/s^2.$$
The net force acting on me includes force of gravity \(F_g\) and the floor pushing me up \(N\) in response to gravity and the force I am applying to the floor to initiate the jump.
$$F_{NET} = N + F_g$$
The force of the floor pushing me up is in response to the force I am applying and the force of gravity. So, \(N = -F_g - F_a\). The result is that the net force on my CoG is the force I am applying (in the opposite direction; I am pushing my feet down on the floor but the net force is upward).
(When it comes to the mass involved there is a bit of a conundrum. I am currently about 225 lbs. But the force I generate as part of the jump is not acting to directly accelerate all 225 lbs. Certainly the floor is pushing back on all 225lbs of me though. From the perspective of the force I am applying, I could probably exclude the lower leg. I found an estimate for men of about 4.5% per lower leg. So, knock off 20 lbs. Then the mass I am accelerating is about 205 lbs. However, we started down this path talking about the CoG and we included the lower leg in this. Also, we are thinking in terms of external forces in our basic force equation. If we want to get super detailed, we will need to think of internal forces and that can get complicated. In the end, the calculation appears only to be affected by fractions of an inch. I'm hunting for elephants so I will not make a careful search of the short grass.)
Using my full weight in kg, let's see what my acceleration would be if I lost 10kg. I'm currently about 102kg. Then the projected acceleration can be calculated from
$$F_a = m_i a_i = m_f * a_f$$
$$a_f = \frac{m_i a_i}{m_f} = \frac{102 kg \cdot 11.1 m/s^2}{92 kg} = 12.3 m/s^2$$
Now we can get my new take off velocity:
$$v_f = \sqrt{2 d a} = \sqrt{2 \cdot 0.38 m \cdot 12.3 m/s^2} = 3.06 m/s$$
2. The Easy Part
My new vertical, D, is an easy calculation with our new take off velocity.$$D = -\frac{v_f^2}{2 g} = -\frac{(3.06 m/s)^2}{2 \cdot -9.81 m/s^2} = 0.477m = 18.8"$$
So, if I lose about 22lbs, I will have almost a 2" higher vertical without developing any further explosive strength in my legs.
Maybe I should try other strategies.