Till I pursued a geometric analysis off late, I had been having a serious doubt that the work done formula for linear motions (W= F x D) may not suit the work done by rotational motions occurring in our joints.
In this chapter, I provide logical reasoning that the work done formula for linear motions can also be used to calculate the work done by rotational motions.
To understand this I did the following experiment: Step-1: Draw a line 10 cm long.
Let the left end of this line be 'A' and the opposite end be 'R'.
Plot a point 'E' 3 cm away from point 'A' (between A & R).
Draw another line from point 'E' 45 degrees angled towards the left side of the line AR.
Let the opposite end of this new line be O (so line EO is also drawn whose length must be 9 cm.
Now consider, A- Axis of a joint, R - Resistance or load, Distance between A and R as Resistance arm (RA), E - Effort or point where the muscle takes effort, Distance between A and E as effort arm (EA), O - origin of the muscle, Line AR is a bony lever (example: Forearm) and the portion from O to E as a muscle.
Let us also assume 1 cm as 1 m in our experiment.
Step-2: Imagine 10 N is applied at point R.
Under this circumstance, the entire lever can tend to rotate downward around the axis 'A' through gravitational torque.
The magnitude of this gravitational torque (GT) will be; GT = R x RA = 10 N x 10 m = 100 Newton meters To counter-balance this GT, anti-gravitational torque (AGT) must be produced at point E using the available effort arm (3 m).
AGT = (R x RA)/ EA = (10 N x 10 m)/ 3 m = 33.
3 N But, 33.
3 N must be applied at 90 degrees to the line AR at point E.
According to our experiment, the action line of muscle is 45 degrees, so more than 33.
3 N must be applied in this angle to get 33.
3 N acting perpendicular to the line AR at point E.
The resultant force (RF) acting at 45 degrees on point E can be calculated as; RF = (R x RA)/ (EA x Sine Θ) Where, Θ - Angle at which muscle acts on the bony lever RF = (10 N x 10 m)/ 3 m x Sine 45 degrees = 46 N Step-3: Imagine that the lever has been rotated to about 60 degrees.
You can learn this by drawing another line AR (10 cm long) from point 'A' deviated 60 degrees from horizontal R.
Under this circumstance, the entire lever can still tend to rotate downward around the axis 'A' through gravitational torque.
The magnitude of this gravitational torque (GT) will be; GT = R x RA = 10 N x (10 m x Cosine Θ) = 10 N x (10 m x Cosine 60 degrees) = 50 Newton meters To counter-balance this GT, anti-gravitational torque (AGT) must be produced at point E using the available effort arm (3 m).
AGT = (R x RA)/ EA = (10 N x {10 m x Cosine 60 degrees})/ 3 m = 16.
6 N But, in this situation also 16.
6 N must be applied at 90 degrees to the line AR at point E.
According to our experiment, the action line of muscle would change into 100 degrees, so more than 16.
6 N must be applied in this angle to get 16.
6 N acting perpendicular to the line AR at point E.
The resultant force (RF) acting at 100 degrees on point E can be calculated as; RF = (R x RA)/ (EA x Cosine Θ) Θ - Angle at which muscle acts on the bony lever.
Here convert Sine 100 degrees as Cosine 10 degrees.
RF = 50 Newton meters / 3 m x Cosine 10 degrees = 17 N Step-4: Work done at point R and point E should be calculated.
(a) Work done at point R = R x (RA x Sine 60 degrees) = 86 joules Here 'R' should be the average of what it was at neutral state and what it is after 60 degrees of rotation of lever.
In both the situations, 'R' remains constant, so (10 N + 10 N) /2 gives 10 N.
(b) Work done at point E = E x (EA x Sine 60 degrees) = 82 joules Here 'E' should be the average of what it was at neutral state and what it is after 60 degrees of rotation of lever.
So, (46 N + 17 N) /2 gives 31.
5 N.
If one goes precisely with strict trigonometric values and diagrams, work done at point R and E would equal each other (or work done at E > work done at R).
Finally to calculate the work done during strength training (1) Either measure the vertical height of the starting and finishing point of the moved object - like Dumbbell, Barbell etc.
, (2) or Measure the distance between the starting and finishing point of the weights attached in the machine - like leg press machine, compound row etc.
, (3) Convert the resistance 'R' being moved, into Newton, i.
e, using equation F = ma.
(4) consider one repetition as factor '2', to take both concentric and eccentric phases because work is done during both these phases during each repetition.
For example, a man lifts 15 kg (150 N) barbell during bench press for 15 repetitions and during each repetition the barbell moves up to 20 cm (0.
2 m) neutral.
Then the total work done can be calculated as follows; Total work done = 150 N x 0.
2 m x 30 repetitions (Remember: Each repetition has to be multiplied by 2) = 900 joules.
(Converting this value into calories doesn't give much clarity and thus a detailed research may be required to correlate this value with calories burned through exercise) The total work done might change little to a higher side because the work done during eccentric phase has been opined to be greater than the work done during concentric phase.