Pressure

Pressure is something we use in all our dancing and usually COMes from two sources:

• connection
• standing on the floor

What is 'pressure'?

In simple terms it's the force experienced by an object when it touches another object moving at a slower velocity.

Example: put a book on a shelf and adjacent to the end of the shelf. The book will experience an upward force equal to gravity pulling it down into the shelf. If the book is leaning against the wall it will experience a force proportionate to the amount of mass 'pushing' against the wall.

It's important to understand that when we are standing on one foot, that foot is experiencing an upward pressure from the floor counteracting the pull of gravity. There is no way that single foot can produce any more 'pressure' as it's all down to gravity. Gravity is pulling the foot down and the floor is pushing back against it. Your foot is trying to move faster than the floor and that's where the pressure COMes from.

If we remove gravity, pressure can be very simplistically expressed as the result of muscles pushing against something and in Dancing we do not want to push anything! There is however an inevitable exception... "push into the floor"

Example, pushing into the floor.

This is inevitably interpreted as 'putting weight onto the "pushing" foot' which results in split weight between the feet.

A better example! Stand on your left leg and bend the knee until your right leg can extend to the side with the inside edge of the front of the foot is touching the floor (without weight). It probably feels pretty unstable and your inclination might be to put some weight on your right foot. Don't put weight on your right foot!! Instead try to pull your right thigh towards your standing (left) thigh in a squeezing motion without moving your left leg. This position is a lot more stable as the right leg is pulling against the floor effectively putting tension between the legs and feet. This is the position the Leader would have in an Oversway.

So how much tension?

It depends on geometry and intent. The free leg acts like a rigid bar from hip to floor.
When you “squeeze” the thighs together, you create an adduction torque at the hip.
This torque must be balanced by equal-and-opposite forces at the feet:

• A horizontal shear at the standing foot (left)
• A matching shear at the free foot’s inside edge (right)

No actual weight has transferred — the vertical support is still 100% on the left foot —
but the right foot now feels “pressure” because the leg is pulling against the floor.

How to think of it

• Small tension (gentle squeeze): the right foot just kisses the floor, like a stabilizer.
• Moderate tension: the shear is enough that if the floor were slippery, the right foot would slide inward.
• Excessive tension: you’ll feel the inside of the right foot “bite” hard — not necessary, and often counterproductive.

The exact number of Newtons doesn’t matter. What matters is the ratio: enough to stabilize the frame (as in an Oversway), but not so much that the right leg “fights” the standing leg.

👉 “Pressure is not weight. It’s the tension your legs create when one leg pushes against the floor to stabilize the other.”

A worked example!

The model (frontal plane, rigid free leg)

• Hip height \(H\) (fixed).

• Free leg length \(\ell\) (held straight).

• Foot contact at \(x\) on the floor \(y = 0\) with   \(x^2+h^2=\ell^2\).

• Leg angle from vertical \(\alpha=\arctan(\dfrac{x}{h})\)

When you adduct/squeeze the free leg, you create a hip adduction torque \(\tau\). At the foot, the ground supplies:

• horizontal shear \(F_x\)​ (what you feel as “pressure”/bite),

• normal reaction \(N\) (contact load).

From the statics of this simple setup:

  \($ \boxed{\;F_x=\dfrac{\tau}{h}\;},\qquad \boxed{\;N=\dfrac{\tau}{x}\;},\qquad \boxed{\;\dfrac{|F_x|}{N}=\dfrac{x}{h}=\tan\alpha\;} \)$

The friction cone (no slip) requires:

  \(\boxed{\tan α≤μs}\) (otherwise the inside edge must slide/roll).

These three boxes are the whole proof engine: geometry:

• sets \(\tan\alpha\);
• \(\tau\) sets how big \(F_x\)​ and \(N\) are;
• sticking is possible only if the geometry fits the floor’s \(\mu_s\).

Why “pressure \(\ne\) weight”

“Weight transfer” would use friction generated by vertical load w mgw\,m gwmg on the free foot:

\(|F_x| \le \mu_s \,(w\,m g)\)

To produce a target torque \(\tau\) using weight alone:

\(\dfrac{\tau}{h} \le \mu_s\, w\,m g \quad\Rightarrow\quad \boxed{\;w \;\ge\; \dfrac{\tau}{\mu_s\,m g\,h}\;}\)​​

This says exactly how much body-weight fraction you must shove onto the free foot to get the same shear.

But with the rigid-leg squeeze, you don’t need to “shift weight” to create shear; the leg’s internal force transmits a constraint reaction that produces \(N=\dfrac{τ}{x}\) and \(F_x=\dfrac{\tau}{h}\) even if your overall vertical support remains on the standing foot. In other words, the scale under the free foot can read near-zero body weight, but still show a nonzero contact force due to the bar-like leg.

When is squeezing strictly better than “use weight”?

1. Aesthetics / balance: weight transfer shifts your COM and base of support; squeeze doesn’t (or far less).

2. Control: with weight, \(F_x\)​ rises and falls with \(w\); with squeeze, you dial \(\tau\) directly.

3. Feasibility: if the geometry violates the friction cone (\(\tan\alpha>\mu s\)​), no amount of extra press will make it stick—you must change geometry (tiny rise ↑\(H\), bring foot in ↓\(x\), or allow a glide).

   Concrete numbers (so it's not just opinion)

For example: \(h=0.90 m\),  \(x=0.45m → \tan\alpha=0.5\) (works for many floors with \(\mu_s\approx 0.6μs\)).
Pick a modest stabilizing torque target: \(\tau=50N\text·m\) at the hip.

Squeeze method (rigid free leg)

\(F_x=\dfrac{50}{0.90}\approx 55.6\text{ N},\quad N=\dfrac{50}{0.45}\approx 111.1\text{ N},\quad \dfrac{F_x}{N}=0.5 \ (\checkmark\; \mu_s=0.6)\)

You get the shear you want, with minimal COM shift.

Weight method
Let \(m=70\text{ kg},\; \mu s=0.6\)

\(w \ge \dfrac{50}{0.6\cdot 70\cdot 9.81 \cdot 0.90} \approx \dfrac{50}{370.4}\approx 0.135 \;=\;13.5\%\)

You’d need to shift ~14% of your body weight onto the free foot to get the same shear from friction alone—visibly changing balance and often the look of the figure.

Proof punchline: For the same stabilizing torque \(\tau\), squeezing can deliver the needed shear with near-zero weight transfer, whereas weight-based friction needs \(w\ge \dfrac{\tau}{\mu_s m g h}\)) and moves your COM.

What if your current geometry is “sticky” or impossible?

Using your earlier numbers \(h=2.5,\; x\approx 3.12\Rightarrow \tan\alpha\approx 1.25\) Even a grippy floor with \(μs=0.8\) fails the cone: \(1.25>0.8\) ⇒ sticking impossible.
You must change geometry (slight rise: ↑\(H\), or bring foot in: ↓\(x\)) or accept a controlled glide (kinetic friction).

A 2-minute, at-home experiment (proof you can feel)

• Put two bathroom scales side-by-side.

• Stand fully on the left scale; right foot inside edge on the right scale (reading \(≈ 0–5lb\)).

• Loop a luggage scale/strap around a fixed post at hip level to your right ankle (measures horizontal pull).

• Now squeeze the right thigh inward without moving the left.

  • You’ll see the luggage scale show a horizontal force \(F_x\)​ while the right scale shows little weight.
  • If you try the same shear by adding weight instead, watch how the right scale reading must rise (and your COM shifts).

This demonstrates “pressure is not weight” and matches the equations above.

• Pressure is the contact force your free leg can create against the floor via squeeze, without taking weight.