# Why knee and hip torques in Squat are always referred to the midfoot line?

The knee and hip moment arms are, in Squat, always evaluated with respect to the midfoot line.

This choice is not obvious for me. In fact, the midfoot is just the point where the centre of gravity of the system Athlete + Barbell must be. But the knee and hip torques have nothing to do with balance. As shown in this topic, these two torques are already balanced, and the purpose of a diagram such that shown above is just to show which joint is under more stress in a certain position (in our case the bottom of a barbell back squat).

So: why do we choose the midfoot line as a reference to say if the knee or the hip is under more stress?

To simplify matters let us ignore the weight of the body. In this case the barbell must be placed directly over the pressure point of the groundforces on the foot. The groundforces are in fact distributed over the feet with varying pressure (red: high pressure):

However in a mechanical analysis these distributed forces may be replaced by a single force applied in the pressure weighted centre of the foot (black in figure): the pressure point.

Further let us model the lower leg as a fixed rod, T, and the upper leg as a fixed rod, F. Let us model the knee extensors as an electrical engine placed inside the knee opening up the knee angle with a moment Mk and the hip extensors as an electrical engine placed inside the hip opening up the hip angle with a moment Mh:

First let us analyze the moments working on the knee. Considering the static case the sum of these must be 0. Which means that: Mk = Fg * dk + Fb * dk But in the static case Fg = Fb = F =>

Mk = 2 * F * dk

Next let us analyze the moments working on the hip. Considering the static case the sum of these must be 0. Which means that:

Mh = 2 * F * dh

As we see from these equations:

• the hip moment is proportional to the horizontal distance from the hip to the pressure point on the foot

• the knee moment is proportional to the horizontal distance from the knee to the pressure point on the foot

And the pressure point on the foot is roughly midfoot.

The above is a first order approximation. The hamstrings cross both the hip and the knee. They therefore produce both hip extension torque and knee flexion torque. Likewise the rectus femoris cross both the knee and the hip. They therefore produce both knee extension torque and hip flexion torque. However due to where the insertions are placed when the hamstrings contract, the amount of hip extension torque they produce is considerably greater than the amount of knee flexion torque they produce. The opposite is true for the rectus femoris: It produces much more knee extension torque than hip flexion torque. (1)

• Calculating the moment arm from the ankle joint to the knee gives you the net moment acting on the shin bones. This includes the moments created by the calf muscles at the ankle as well as the moments at the knee created by the quadriceps and hamstrings. Whereas a moment arm from the centre of pressure to the knee gives you only the moment acting at the knee. Jan 1, 2022 at 12:32
• Andy and @DavidScarlett how would you define "the centre of pressure"? Why is it exactly the midfoot? Jan 1, 2022 at 17:40
• @Kinka-Byo: it depends on your technique. It may for instance be slightly more towards the heel in which case you have the weight more "on your heels". However this horizontal shift from the midfoot is small compared to the knee and hip moment arm and may be ignored in this first order analysis.
– Andy
Jan 1, 2022 at 19:01
• The actual physics here seem rather dubious. (1) Hips and knees are joints, they don't have plain/regular "momentum". Is this angular momentum? Angular momentum is the moment of inertia multiplied by the angular velocity. Where is that in your answer? (2) Torque is only the product of moment-arm and weight if you assume everything else is "completely-rigid" outside of the angle (of interest) you're computing the torque at. But this is clearly not the case for any multi-joint exercise such the squat (you technically have three "non-rigid" angles here -- the hip, knee, as well as the ankle). Oct 10, 2023 at 10:12
• You can however always compute the exact torque at an angle A indirectly by measuring `dW/dA` -- the amount of work W done (in this case, as a change in the potential energy of the weight by lifting it up) for a given change in that angle A. This can be computed for a two-joint/angle exercise like pull-ups easily (shoulder and elbow angle), but for a three-joint exercise like the squat (hips, knee, and ankle angle) there are an infinite number of angle configurations for a given barbell height, so this is not as straightforward there. Oct 10, 2023 at 10:12