# Physical Model of Exercise

Trying to model caloric effects on the body when exercising. I think I have something for lifting weights, but at a loss for running or bicycling.

1. Lifting Weights. I think this is pretty easy. Raising a $$x$$ lb bar $$y$$ ft takes $$xy$$ ft-lbs of energy. Lowering it slowly also takes energy, the previous quantity, minus kinetic energy at end of drop, just before impact at bottom. So best case scenario for a $$50$$lb weight lifted abd lowered $$2$$ ft, that's 0.064 Calories or kilocalories per lift cycle.

2. Suppose you are walking at constant speed $$v_0$$ for time $$\tau$$. What's the energy output from your body? I'm thinking $$E=v_0\tau F_{avg}$$ where $$F_{avg}$$ is the average of the net forces working against the motion. Are there other forces acting besides wind resistance and friction? I'm thinking wind resistance is $$F_w=-bv_0$$ for appropriate $$b$$. Friction is $$F_f=w\mu$$ where $$w$$ is weight and $$\mu$$ is co-efficient of static force.

Typical values are $$v_0=4$$ mph, $$\tau=$$ 3600 s, $$E=350$$kcal$$=1.08e^6$$ftlb $$\implies F_{avg}=\frac{E}{v_0\tau}=51.13$$lbs.

For vertical forces in the running/walking scenario, just remaining erect is not letting gravity drop you to the floor. So you input as much as gravity takes per second. $$\frac{wh}{2}\sqrt{\frac{g}{h}}=w\sqrt{\frac{gh}{4}}$$ is calories burned just from standing up. That's weight falling from center of mass to the ground divided by the time of free fall for an object at that height, half of $$h$$, the height of the person in question.

If you could maintain a lift every two seconds for an hour, the calorie burn would be over 100 times that due to running, right? Two seconds an hour is 1800 lifts burning 0.064 kcal each.

Are my forces at all accurate for the walking scenario? Does weight of the person only factor in for friction and standing?

• Exercise physiology text books should have something to say about this subject. Jan 14, 2022 at 20:03