For a medical experiment I am doing, I need an equation to find the angle at which someone will lean before falling. I am not mathematically inclined in terms of advanced stuff, I am more so of a trigonometry person.

I assume factors that will be needed are BMI, including height and weight, but that is really all I have. For example, I am a 5'11" (180.3cm) tall female. I weigh 165 pounds (74.8kg). I want to find out, computationally (because I could measure myself), how far I could lean before I fall.

Any ideas of how I could go upon this?

  • $\begingroup$ @ probably_someone Won't they fall if the cg is basically no longer above the footprint? $\endgroup$
    – Tom B.
    Mar 3, 2018 at 2:22
  • $\begingroup$ @ probably_someone Well, if they spread the feet, and lean forward or backward, then the circle introduces even more error, no? I just don't see the need for the circle, when the footprint does the job exactly. Granted, we're dealing with a very simplified case due to the nature of the question, but nonetheless. $\endgroup$
    – Tom B.
    Mar 3, 2018 at 2:30
  • $\begingroup$ You're correct; spreading the legs gives you no forward-backward advantage. It should really be the smallest rectangle containing both feet. $\endgroup$ Mar 3, 2018 at 2:34
  • $\begingroup$ @probably_someone: If we are really making differences between circle/rectangle, we might as well state that the center of gravity may not be outside the convex hull around both footprints... But even then: Feet and body are not rigid, and the question does not even define whether the arms must be held alongside the body, or if they may be used as counter weights... $\endgroup$ Mar 3, 2018 at 3:21
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    $\begingroup$ To illustrate the problem of finding such an angle because you always try and compensate for a lean try the following. Stand against a wall with no skirting board with back of your feet and your buttocks against the wall at all times and try and pick up an object on the floor in front of you; you will faito pick up the objectl? This illustrates the fact that when you lean forward to maintain balance you tend to move you buttocks in the opposite direction. $\endgroup$
    – Farcher
    Mar 3, 2018 at 8:42

2 Answers 2


Even though your body is not a simple, homogeneous, rigid object, we can calculate a little... With a lot of assumptions.

I assume, your feet stand next to each other, not in a step-forward-position. And I assume, the question is about leaning forward/backward, not sideways. And I assume that the arms have to be aligned at the side of the body.

First step, we need to locate your center of gravity (CG). Even if you feel light/heavy hearted, we assume you to be symmetric. We also assume you to be front-back-symmetric, except for your feet.

Now we need the hight-coordinate of CG. You can find it via plancking across a bar. Let's assume your result is that CG is 99cm from floor up.

Next up, we need your shoe size. I guess you at an american women's size of 11.5, which equates to 27cm foot-length.

If you ever took dance classes, you may remember that you were instructed to "put your weight over your heel", or over the center of your feet. Let's assume that this little shift can be done without "leaning".

As you are "more of a trigonometry person", here comes the fun part:

We think of you (having "weight over heel") as a L-shape, with your feet being 27cm, and the stem of the "L" being 99cm (height of CG). Now you gradually lean forward, and when CG passes over your toes, you fall. This will happen at an angle of $$\alpha_{max} = arcsin \Big( \frac{27}{99} \Big) = 15.8° $$

With the weight over the center of your feet, the "usable" length of your feet halves. So, we arrive at a maximum leaning angle of $$\alpha_{max} = arcsin \Big( \frac{13.5}{99} \Big) = 7.8°$$

This was all a lot of assumptions. But it got us somewhere. Now, we can consider a maximum leaning angle of 8°-15° as somewhat realistic.

And, by the way, we just went for the geometry of your body - your weight is not part of the equation.

  • $\begingroup$ I really was not asking for specifics or a 100% accurate result, so this is way more than perfect for what I would need. I really need my results to be accurate within a few radians, so I think your model would work perfectly. Time will tell, but you are scratched in my notebook for now. I appreciate your time and far superior knowledge :) $\endgroup$ Mar 3, 2018 at 3:56
  • $\begingroup$ @Autumn: It was fun to play around with your problem! As a small hint for your next question: Maybe be a bit more specific (e.g. what are the arms doing, leaning forward or side ways...). And keep your units in order: Our result is 8-15 degree. One radian is ~58 degrees. Leaning "a few radians" forward will get you a somersault ;-) $\endgroup$ Mar 3, 2018 at 4:07
  • $\begingroup$ Oops, it was late for me last night when I said radians urghhhhh this is why you sleep early, children. Plus, with specificity, it is a bit chaotic. My experiment will involve cerebral palsy, so the subjects are not expected to fall a certain which way. My next thought is to do this model for every possible way a person could fall in a 4 way Cartesian fashion (forward backward left right) and average it out. Thanks a ton! $\endgroup$ Mar 3, 2018 at 14:23
  • $\begingroup$ @Autumn Well, your problem indeed seems a bit chaotic. In the calculation, we assumed the person to stiffly lean forward, arms at the sides, no further (compensational) movements. There are dozens of ways to fall down, without meeting our specific criteria. E.g. standing, then suddenly relaxing all leg muscles: Knees and hips give way, and before your upper torso is even near a 10° leaning angle, you already plunged about a feet and a half into an unrecoverable position. But I have to admitt: More physicist than physician, I have no idea, what implication cerebral palsy may carry. $\endgroup$ Mar 4, 2018 at 1:27

In normal rigid-body mechanics, an object tips over if its center of gravity passes beyond its tipping point. The tipping point for a human is roughly the outer edge of the footprint. Assuming the feet can adopt any configuration, the effective "outer edge of the footprint" is the smallest rectangle that includes both of the feet. When the person's center of gravity goes outside this rectangle, they are likely to fall over.


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