Kinetic Energy and the Human Body - Shot Put Illustration

In a previous post (Remembering Gravity), I was estimating the change to my vertical jump if my strength, speed, & power capabilities remained constant and my weight decreased. I mentioned that one of my assumptions was that the amount of change in the weight involved was probably such that the force production of my legs was the same on the light me as the heavy me. I also suggested that if the amount of difference in weight was significant, this assumption would start to unravel. My example to motivate this at an intuitive level was to compare shot putting with a 12 lb shot with shot putting a softball. Intuitively, you can probably recognize, that producing the same "power" with the softball as with the shot would involve an incredible amount of speed and that the result of shot putting a softball would be governed by maximum speed production rather than maximum force production. I thought it would be interesting to put numbers to this illustration.

When I jump or throw or "put", I am imparting force over distance. That force may not be constant and the direction may not be either but at the end of force application I have imparted a net amount of work to an object (my own body, ball, shot). In the end, the object has an amount of kinetic energy
\[W_{NET} = K = \frac{1}{2} m v^2.\]
So, to state my illustration more technically, I am saying that, using an identical movement pattern, allowing only changes in speed of movement of my body, I cannot impart as much kinetic energy to a very light object as I can to a heavier, but still manageable object.

So, to establish the general invalidity of the assumption of equivalent force production under different loading we can put a 7.260 kg shot and a 190 g softball (or 146 g baseball) and determine the approximate amount of kinetic energy imparted. A little hitch would be the need to know the release angle. An interesting post on Brunel University's site (Shot Put Projection Angle) highlights that in practice the optimum release angle is not 45° but may be between 26° and 38° and and this is due to a biomechanical advantage favoring superior speed production in the horizontal direction.

Let's Pretend

Suppose we take a world record shot put of 23.12 m with a 7.260 kg shot, assume a 30° release angle and determine the kinetic energy imparted to the shot by the putter. Then we will make the very erroneous assumption that the same person would have been capable of imparting the same kinetic energy to a softball or baseball, using the same biomechanical movement pattern. We are expecting to get a ridiculous answer which will satisfy us that the assumption that we can do this is bogus.

To determine the kinetic energy of the shot, I need to find the initial velocity. We use a few common kinematic equations. We can use the relationship between the vertical and horizontal velocity which is determined by the release angle, so that \(\tan \theta = v_y / v_x\), where \(\theta\) is the release angle. We neglect the effect of drag from the air. Taking \(X\) as the horizontal distance traveled by the shot, we have
\[X = v_x t_t = \frac{v_y}{\tan \theta} t_t\]
and so
\[t_t = \tan \theta \frac{X}{v_y}.\]
Using the vertical displacement equation, we have
\[0 = d_R + v_y t_t + \frac{1}{2} g t_t^2 \]
\[0 = d_R + v_y \Big(\frac{\tan \theta}{v_y} X\Big) + \frac{1}{2} g \Big(\frac{\tan \theta}{v_y} X\Big)^2\]
\[0 = d_R + X \tan \theta + \frac{g X^2 \tan^2 \theta}{2 v_y^2} \]
\[\frac{X^2 g \tan^2 \theta}{2 v_y^2} = -d_R - X \tan \theta\]
\[2 v_y^2 = - \frac{X^2 g \tan^2 \theta}{d_R + X \tan \theta}\]
\[v_y = X \tan \theta \sqrt{\frac{-g}{2 (d_R + X \tan \theta)}}\]
We have our velocity, \(v_p\) of the shot put, then as \[v_p = \frac{v_y}{\cos \theta}.\]
Here's where we make our faulty assumption. (Faulty, because we are comparing objects of very different masses.) We take the kinetic energy the putter could generate on the shot and assume he can generate the same kinetic energy with a much smaller mass. Hence,
\[\frac{1}{2} m_p v_p^2 = \frac{1}{2} m_s v_s^2 \]
and so
\[v_s = v_p \sqrt{\frac{m_p}{m_s}}.\]
Since we are assuming biomechanical identity, we use the same release angle giving us the vertical initial velocity of the softball, \(v_{sy}\)
\[v_{sy} = v_s \sin \theta \sqrt{\frac{m_p}{m_s}}.\]
and horizontal
\[v_{sx} = v_s \cos \theta \sqrt{\frac{m_p}{m_s}}.\]
From here we proceed to determine how long the ball will be in the air. There will be a time to the peak \(t_p\) and a time to drop \(t_d\) in the total time \(t_t\). For the time to peak we have
\[0 = v_{sy} + g t_p\]
\[t_p = -\frac{v_{sy}}{g}\]
From this we get our maximum height, \(h_p\)
\[h_p = d_R + v_{sy} t_p + \frac{1}{2} g t_p^2\]
\[= d_R - \frac{v_{sy}^2}{2 g}\]
This allows us to find our time to drop from the displacement equation as
\[0 = h_p + \frac{1}{2} g t_d^2\]
\[t_d = \sqrt{\frac{2 h_p}{-g}}\]
So, we can give an expression of our bogus displacement of the softball as a shot put by
\[X_{sb} = v_{sx} (t_p + t_d).\]
We can follow through these expressions to give output without spending the time to make a monster expression out of it. I looked at doing that briefly and it didn't look like there was much for cancellations that made the expression more tidy. So, there is no computation savings by combining the expression into one monster formula. Therefore, since I want a number, I will just put these expressions into maxima and crunch.

Here's the sequence of expressions in maxima:

theta: float(30 * %pi/180);
m_p: 7.260;
m_s: 0.190;
d_R: 2.0;
X_sp: 23.12;
g: -9.81;

v_y: X_sp * tan(theta) * sqrt(-g/(2*(d_R+X_sp*tan(theta))));
v_p: v_y / cos(theta);

v_s: v_p * sqrt(m_p/m_s);
v_sx: v_s * cos(theta);
v_sy: v_s * sin(theta);
t_p: -v_sy/g;
h_p: d_R - v_sy^2/(2*g);
t_d: sqrt(2*h_p/(-g));

X_sb: v_sx* (t_p + t_d);

The final answer comes out at about 259.52 m for a softball. Changing the mass to a baseball, gives 336.71 m.

Athletes who use biomechanically superior form to throw a baseball actually come in more about 136 m or so (Glen Gorbous).

So, if you're trying to use the human body to generate kinetic energy, there's a mass sweet spot somewhere. Too little mass and you just can't generate the speed. Too much mass and you can't move it at all. Somewhere in between is the optimum, if, for reasons unknown, you care about generating a lot of kinetic energy.

Remembering Gravity

I was recently testing out my vertical and was disappointed in the results. I was also mindful of my current endeavour to slim down a bit. And the thought came to mind, If I lost 20lbs, what would the impact be on my vertical jump? My intuition at that moment was that it should easily be possible to calculate an estimate. But as I thought about it some more, I realized I had oversimplified the matter and I remembered how gravity works.

There are two main pieces to the puzzle:

1. Determine the effect on take off velocity.
2. 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:

1. Near constant vertical acceleration.
2. 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.