Calculating Rotational Force on Knee Joints for Karate Stances

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SUMMARY

This discussion focuses on calculating the stress on knee joints during low karate stances, specifically for a 200-pound individual with an 18-inch shin and a 24-inch thigh. The user seeks to determine the force on the knee joint at various angles: vertical, 45 degrees, 90 degrees, and 105 degrees. Key insights include that each knee supports 100 pounds when stationary, plus additional tension from muscles. The complexity of the model is acknowledged, with considerations for the pivot point and ligament attachment affecting stress calculations.

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matb
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I'm trying to calculate the stress placed upon knee joints by low karate stances.

If I assume a body weight of 200 pounds, with a shin length of 18 inches and a thigh length of 24 inches, how can I calculate the force placed upon the knee joint when the thigh is:

vertical above the shin
at 45 degrees to the shin
at 90 degrees to the shin
at 105 degrees to the shin

Please provide formulae so that I can apply to other situations.

Thanks very much in advance
 
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If the 200lb person is not moving, and the legs are equi-distant from the center of mass, then each knee supports 100lbs, plus any tension force required to hold the position. Surface area of the knee joint contact point is a factor, as well as the positioning of the connection points for ligaments; these factors vary a bit from person to person.

The information I don't have is the distance from the effective pivot point of the knee to the point of attachment for the ligaments that are in turn attached to muscles that are supplying a supportive force to hold the position.
 
Oh my goodness! I didn't realize it was that complicated.

Is it not possible to use a greatly simplfied model that assumes a simple hinge joint and evenly distributed connecting tissues spread across its width?

In the first instance, I'm simply trying to work out the increase in stress on the joint as the angle of the thigh increases. I read somewhere that it was equivalent to three times the normal body weight at 90 degrees, but I was hoping to get the maths to prove it...
 

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