2
$\begingroup$

When the human body is perfectly centered, the center of mass is on the centerline around the belly button. However, the human body is not always in this rigid posture. We have many postures that we use. How much does the center of mass shift, left to right, if we look at different stances. I'm not so much interested in dynamic motions like walking, but rather static positions like:

Obviously the center of mass will shift with respect to the ground, staying between the feet, but what I'm interested in is how it shifts with respect to the torso's centerline.

$\endgroup$
  • $\begingroup$ I'm voting to close this question as off-topic because although the question involves living things, it's not a biological question. This is a physics/mechanical engineering question. $\endgroup$ – James Aug 23 '16 at 4:42
  • $\begingroup$ @James While the calculation is pure physics, the numbers that would be needed can only be determined empirically based on how human beings carry themselves. The only source I can think of to acquire such numbers would be biology. $\endgroup$ – Cort Ammon Aug 23 '16 at 5:08
  • $\begingroup$ This question is the field of biomechanics (video here). There is of course overlap. However, the simulations and measurements carried out in order to answer this question will be way more at the mechanical engineering end of the spectrum. $\endgroup$ – James Aug 23 '16 at 6:05
1
$\begingroup$

You can find the relevant math here:

http://hypertextbook.com/facts/2006/centerofmass.shtml

The key factors are the body mass distribution, height, standing center of mass, heartbeat, clothing, etc. You can approximate the center of mass for a given stance as:

In order to determine the center of mass of a person, we used the formula for torque:

$\Sigma\tau = r \times F$

The torques due to the weight on each end of the person were set equal to each other. The displayed weight on the scale was the force, and there are two different radii. The first radius is the distance to the center of mass from the person's feet, and the second radius is the length of the board minus the distance to the center of mass from the person's feet. The net torque of the system is zero and therefore the torques on the opposite sides of the boards must be equal:

$w_1\times x = w_2\times (l~-~x)$

where w1 is equal to the weight at the person's feet, x is equal to the distance from the person's feet to his/her center of mass, w2is equal to the weight at the head, and l is equal to the length of the beams. The resulting formula, when solved for the distance to the center of mass from the person's feet (radius one) is:

$x = \underline {~~~w_2\times l~~~}$

$\hspace {11mm}w_1~+~w_2$

After determining the location of each person's center of mass, the ratio of the center of mass to the height of each person was calculated using the formula:

$x/h$

where x is the location of the person's center of mass and h is the person's height.

| improve this answer | |
$\endgroup$
  • $\begingroup$ That article had plenty of discussion of the center of mass from head to foot, but no content on left to right. Part of this is because the study was done laying down, rather than in a standing position. $\endgroup$ – Cort Ammon Aug 23 '16 at 19:27
  • $\begingroup$ Its the same math. Just treat right as "head" and left as "foot". The fundamental problem is balancing two loads around a center of mass. $\endgroup$ – D J Sims Aug 23 '16 at 19:37
  • $\begingroup$ Its the same math, but different data. I suppose that if nobody has ever done this study before, I can go gather my own data with this method. $\endgroup$ – Cort Ammon Aug 23 '16 at 20:12
  • $\begingroup$ COM moves both horizontally and vertically during gait. However, the displacement is much more higher in up-down than left-right shift $\endgroup$ – bantandor Nov 25 '16 at 11:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.