# Can you remain balanced if your CG passes in front of your toes?

Can you remain balanced if your CG passes in front of your toes? I'm looking at the physics of walking. It's described as a controlled fall for good reason.

What I'm interested in is whether you can recover from the following situation. You have one foot on the ground (let's say it's your left), and one foot (right) is in the air moving forward. You make a decision that you cannot step down with your right foot (perhaps there was a scorpion where you intended to place it). However, at this point in time, you have a forward velocity and your CG happens to already be in front of your left toe. Is there a way to recover such that you can return to a stable balancing position on your left foot without hopping?

I know for a simple rigid body this is impossible. Once the CG is in front of the contact patch, the simple rigid body will tip over. I'm less certain this is true for a human body because we have so many ways to regain our balance. I can't tell if there is a clever way we can move such that we can arrest our forward momentum (probably using friction with the ground) and return the CG to a stable position above the left foot.

• One can of course hop using the left foot. Sep 9, 2016 at 3:07
• @JonCuster Good point. My intent was to not hop. I've edited accordingly. Thanks! Sep 9, 2016 at 3:17
• @JonCuster Does a hop actually stop you rotating forward? Sep 9, 2016 at 8:36
• @Farcher - the hop itself does not. The landing with your foot in front of your CG and the torque it exerts does. You can try it (carefully) at home. Sep 9, 2016 at 12:46
• I did but failed. Sep 9, 2016 at 12:47

## 3 Answers

Yes you can recover from falling without jumping or use of your right foot and you have been doing this ever since you were a kid and started to toddle.

You push you left foot with a sudden force back while at the same time rotating your arms and torso to left quickly. Like a move from swing dance.

What you did, you created a momentum countering the imbalance and bringing your body back to equilibrium.

Many of classical dance steps are actually composed of basic moves: you lean into an exaggerated angle and then gracefully recover and even continue swinging past the balance point to other extreme albeit all synchronized gracefully with music.

Going back to the toddler, he or she is clumsily trying to do the same thing: recover from a miss calculated step by using their arms and wobbling back to balance. Cats are known to be masters if balance.
let's put some numbers in here to give a better perspective. Our walking stride is roughly 16 inches give or take, amazingly most of the course of a step both of our feet are touching the ground except they are transferring our weight from one to the other, running is different though.

Let's say your CG is fallen 3 inches outside of your toes and therefor has created an overturning momentum of:
160lbs x 0.25 ft= 40 ft.lbs.( assuming you are a fit man). now we don't want to get into the thick of calculating our arms' rotational moment of inertia which can be quit involved but we assume half of the mass of our arms concentrated at our fists say 30 inches, and the weigh 9 lbs. each. So we have 9x2= 18/2 = 9 lbs.

the angular momentum we need to counter the over turning momentum is L= mxrxv

plugging the numers: $$40= 9*2.5*v$$ V=1.77 foot/second.
I am no dance expert but the speed required seems reasonably normal and achievable. some young guys by swinging their arms and flexing their torso can do repeated cartwheels and would become Olympic champions!

• This is difficult to visualise. Do you have a link to a video? Sep 10, 2016 at 3:31
• @sammygerbil I edited my answer with a basic numerical example please check it out. I would like to make a primitive numerical model of a runner but I would like to put more hinges and masses in it to make it more natural. Sep 12, 2016 at 20:30
• You are confusing moment of force (= torque, unit $ft.lb$) and angular momentum (unit $ft.lb.s$). These are not the same. Torque = rate of change of angular momentum. To create a counter-torque you have to keep increasing the angular momentum of your arms. Sep 12, 2016 at 20:55
• Right thanks. Let's try the work needed to straighten the body, w =fx = 160. cos(tilt angle)*.25 = 160(.25/3.5, CG of man's height).25 = 2.85 lbs.ft. And the work on the arm bringing the body back up: 2.85 = 1/2 I.omega^2 = 9*2.5^2(V/2.5^2) = 9*V^2 therfor V= 1/3(2.85^.5) =0.56 ft/s . Sep 13, 2016 at 0:32

Isn't this just a matter of conservation of angular momentum? If we are about to fall (either forwards or backwards) we instinctively rotate our arms in the direction in which our body is rotating as it falls, ie "windmill", in order to counter-rotate the body back into the upright position. (Possibly this is what kamran is suggesting.)

The illustration shows the torso (large block) tilted forward to the left and about to topple. The arms (small blocks) are being rotated anti-clockwise to produce a clockwise counter-rotation of the torso.

Note (as in comments to kamran) that the rotation of the arms need to accelerate in order to maintain the counter-torque on the torso, because torque = rate of change of angular momentum. If you are holding something small and heavy in your hands, don't be tempted to throw it away in order to so that your hands are free to grab onto something, or to cushion your fall. Instead, use it to increase angular momentum. (Alternatively you could throw this heavy object forward in order to generate a backward reaction force.)

Another (better?) strategy, recommended by this skating instructor, is to "go low." By collapsing like a deckchair, folding at the waist and knees, you reach down, grab your knees and brace. The physics of this is (I think) that you are reducing your moment of inertia so that you require less torque to counter your falling rotation.

A third possibility is to throw the right leg and arms forward while "collapsing" backwards onto the left foot.

See also this long-winded discussion the physics of balance.

Certainly. By placing your left foot on the ground and pushing hard, you will apply a torque which will counteract the torque produced by gravity. This is probably not practical, but you didn't ask about that.

• If your left foot (toe) is the pivot point, pushing with your foot does not generate counter-torque, since this acts through the pivot. Sep 13, 2016 at 19:02