Dead End Isometric model rising and falling object

Copyright Carl Janssen 2024 September 9

Dead End Isometric model energy expenditure as a rising and falling object

I believe this model as is is worthless or at least worthless without some other modification such as friction or something else to result in a non zero value for a cycle length that approaches zero but I do not know yet what that modification is.  I just wrote this to show me that this is a dead end so that I do not have to calculate why it is a dead end again later

 Some people have claimed that Isometric exercise is equivalent to a muscle shortening and lengthening back and forth a very small distance and averaging out in a position which appears to be a stationary position because the distance moved is so short

An attempt will be made to model isometric exercise as equivalent to the work it takes to repeatedly lift a object up a object with a force resisting gravity then letting it drop back down over again in cycles.  Equations will be done to show how various combinations of force as a function of position or time would result in different amount of energy expenditure per time for time values equal to integer multiples of the cycle duration.  Unfortunately the simple models involving alternating back and forth between a constant force resisting gravity and no force resisting gravity do not appear to work correctly to make positions by taking the limit as the cycle length approaches zero because the equations predict 0 work per time in such a case.  Perhaps such a models could be modified by selecting a correct time value per cycle that is not a limit as time approaches zero, maybe something close to 200 to 300 milliseconds which is close to human reaction time for some tasks or perhaps an even shorter time if the brain and or spine is bypassed and it relies on the nerve tissues near the limb to react in some way to maintain the isometric contraction.  Although my suspicion is that this model simply does not work correctly.

This model would not be accurate because it would involve straight line movements and force instead of rotations and torque but none the less it will be shown.  For very small angles close to 0 radians the percent error when using theta in radians to estimate the sin of theta is very small.  If the change of angle is close to 0 degrees in the joint it might look very close to moving a object in a straight line when the joint is actually moving an object in a rotational path.  This might make the model similar except perhaps multiplied by a constant to reflect the contraction being more difficult the longer one lever arm of the joint is if the other lever arm of the joint is the same.

If you throw a object up at a initial velocity with no force supplied on that object after it is thrown other than gravity then in the absence of friction it's position with respect to time will be symmetrical it will achieve maximum height when it's momentum is zero and it's kinetic energy is zero from a certain reference frame.

A object will be modeled starting stationary then experiencing a constant force resisting gravity for a certain amount of time until it reaches a certain height followed by that force being no more once that height and time has been achieved but the object then moving upward like a projectile experiencing the force of gravity with a certain initial velocity after that time the object will eventually reach a maximum height at a certain time and then be stationary followed by returning to the height at which the upward force resisting gravity stopped, upon achieving that height once more the same upward force will be applied to it again slowing it's fall until it reaches the starting pattern.  This will be repeated in a cyclical pattern.  There will be two different halves two this pattern each forming a local maximum or minimum for the height at which the cycle pattern will be symmetrical with respect to force as a function of height and also force as a function of time.  

The cycle could be divided into four parts 

1 stationary start at bottom until force resisting gravity ceases with object moving up 

2 force resisting gravity ceases and object moves up until it reaches maximum height

3 object moves down until it reaches the height at the end of 1

4 object moves down but experiences the same force in 1 until that force makes it stationary followed by returning to the first part of the cycle

2 is symmetric with 3

1 is symmetric with 4

2 and 3 will be called projectile phases

1 and 4 will be called resistance phases

Nmg = Amount of constant force resisting gravity in height and times that are part of 1 and 4

(N-1) * mg = Amount of net force in heights and times that are part of 1 and 4

- mg = Amount of force in 2 and 3 due to gravity alone

Tres = Resistance time = Amount of time spent in 1 and 4 per cycle

Tpro = Projectile time = Amount of time spent in 2 and 3 per cycle

Tcyc = Amount of time spent in entire cycle = Tres + Tpro

m = mass of object being repeatedly lifted or thrown and dropped and then decelerated

Ignoring direction values and just looking at magnitudes or absolute values

Pmax = Maximum upward momentum object experiences

Pmax = Maximum downward momentum object experiences

Hpro = Change in height from start of phase 2 until end of phase 2

Hpro = Change in height from start of phase 3 until end of phase 3

Hres = Change in height from start of phase 1 until end of phase 1

Hres = Change in height from start of phase 4 until end of phase 4

Pmax = (N-1)gm*(Tres/2)

Tpro / 2= Pmax /gm = (N-1)gm*(Tres/2) / gm

Tpro =  (N-1)*Tres

Pmax = (Tpro/2)*gm

Pmax = (Tres/2)*(N-1)

KEmax = 0.5 * Pmax^2 / m

KEmax = Hpro*g*m = Hres*(N-1)*g*m

Hpro = 0.5*g*(Tpro/2)^2=0.5*g*((N-1)*Tres/2)^2

KEmax = Hpro*gm = Hres * (N-1)*gm

Hres = Hpro / ( N -1 )

PR = Power Requirenent = KEmax / (Tpro +Tres) = KEmax / Tcyc

Tcyc = Tres + (N-1)*Tres = N*Tres

Tcyc = (Tpro) + (Tpro) / (N-1) 

Tres = Tcyc / N

Tpro = Tcyc *(N-1) / N

PR = KEmax / NTres

I do not care to solve it for the exact value today because I believe this model is worthless or at least worthless without adding other factors like friction or something else that gives a non zero final solution and even if I made some mistakes in the exact numbers for the equations about these proportionality should still be true leading to the same conclusion for this model

Final Solution for Power Requirement as cycle length approaches zero

Constant1 is a function of N and m and g but does not change as Tcycle changes

Constant2 is a function of N and m and g but does not change as Tcycle changes

KEMax = Constant1 * Tcyc^2

Power Requirement = Constant2 * Tycl

limit as Tcyc approaches zero of PR equals zero