One error I saw in the document I posted is the experiment they did with loaded depth jumps. They concluded that since athletes did not jump as high with a loaded depth jump it wasn't an effective exercise. Well we know that he ground reaction force will equal the mass of the athlete times gravity. The grf produced from a 3.5 drop landing can be matched with a loaded landing from a lesser drop in an infinite number of ways.
You are confusing absolute force with relative ground reaction force. Read what you wrote. If GRF = mass * gravity.... Then ground reaction force is the same no matter what because mass and gravity are constants.... When you are standing still (or on a scale) the normal force (GRF) is indeed mass * gravity, or your weight in pounds.
The GRF we are interested in is the relative force during the landing (ie when you decelerate from x m/s to 0 m/s - after which GRF = ma again). A force plate can approximate the instantaneous relative force during the landing but this of course depends on the manner of landing. To calculate the average GRF during the landing you can use: averageGRF = (mass * v )/ t + f where v is landing speed and t is landing time.
Now I'm rusty in physics but we do know that distance = (1/2) * g * (t^2) [ one half A T squared ] and that velocity = gt, so we can rearrange and get this formula : V = sqrt ( 2 * d * g )
So take a 100 kg person and have them drop from 0.5 meters. The velocity is 3.1 m/s. If they drop from 1 meter it is 4.4 m/s.
So, the ground reaction force for the first and second case is 310/ t + 980 and 440 / t + 980.
Let's assume our athlete is quick and can jump up again in 0.3 seconds so and about half of that time is spent decelerating to zero, so a landing time of 0.15 seconds.
So our average GRF for each height is: 310/0.15 + 980 and 440/0.15 + 980 or 3046 N and 4109 N. So the average ground reaction force is about 3 times bodyweight or 4 times bodyweight from the two heights if ground contact time is unchanged.
Now assume you only have a 0.5 meter box and you want to get the same GRF as the 1 meter box, so you strap weight onto your subject.
How much mass do we need to add?
I won't bore you solving it but rather say that: ((100+35kg) * 3.1) / 0.15 + 135*9.8 = 4113 Newtons.
So we have to add approximately 35 kilograms to our athlete to recreate the average GRF of a 1 meter drop with a 0.5 meter drop. This is problematic because it might be hard to find a place to sufficiently load an athlete with 77 pounds where he can still have freedom of his limbs and movement AND because the athlete will almost surely not be able to jump up again in 0.15 seconds with 77 pounds of added weight.
Now of course you can add more weight to compensate for the longer ground contacts but I believe the author is making the point that if ground contacts are much much longer the muscle will not be trained in the same weight. Certainly could be so truth to that. One interesting thing to do with these equations is play with force (gravity). If you assume a nice band setup where you can manipulate force so you actually have gravity + bands you can probably maximize the training of the athlete to absorb force while maintaining small GCT.
