The physics of Heels
Entry posted by Momentumous
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So lets say you're feeling crazy and want to wear some heels.
Now you're about 135 pounds, the average weight of a female. That's approximately 61.2 kg. Multiply that by the constant of our friend gravity, and your body exerts a force of about 600N. Granted, this is split up between two feet unless you've had a tragic incident lately, so your foot feels about 300N of force just from standing.
So lets look at it in terms of pressure, psi, pounds per square inch...or rather kg per square inch.
Pressure is equivalent to F/A. If you're shoe size is a women's 8-9, your foot is probably around 10x3 inches. We'll shave off a handful of inches with the assumption that your foot is not perfectly rectangular, so lets say the area feeling the force of your weight is 25 inches.
This means on a normal day with good arch support and nice, flat shoes, each foot feels a pressure of 12 kg/in^2. That's about 23 psi.
So now you think, hey, let's wear some six inch heels!
These heels are tall enough to essentially mean you're constantly standing on your toes. We'll attribute a square inch and a half for your arch "support" and heel, just for kicks, and what's left holding the majority of the force is the front pad of your foot and toes. That's about 8 square inches by my measurements. But the force exerted on your now 9.5 in^2 surface area hasn't changed. The force exerted on this surface area is still 300 N, and this area is less than half of the original surface area. 300/(9.5) is 31 kg/in^2 or 68 psi. That's almost triple the pressure your foot normally feels. That's like a full grown grizzly bear stepping on your toe.
6 inch heels tonight? No thanks, I'll take my nike's.
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