A balance display shows a dot moving to the right. Before describing the person as moving right, identify what that dot represents. It may show a location calculated from contact forces rather than the location of the body's mass. Those quantities answer different questions, even when their traces look similar.
This article develops that distinction through an original mechanical example. The numbers describe an ideal teaching apparatus, not a patient, a balance assessment threshold or a recommendation about treatment. The useful skill is to state what was measured, what was inferred and which assumptions connect the two.
Start with the weights in the average
The centre of mass, or CoM, is the mass-weighted position of a defined system. In one horizontal direction, xCoM = Σ(mᵢxᵢ)/Σmᵢ. A component containing more mass contributes more to the result. The position can lie in empty space: it does not need to coincide with a particular anatomical structure or piece of material. OpenStax: centre of mass.
The centre of pressure, or CoP, describes the resultant contact loading on a specified support surface. For the vertical-force model used below, its horizontal coordinate is force-weighted: xCoP = Σ(Fᵢxᵢ)/ΣFᵢ. A balance experiment may therefore track CoP using force measurements without directly tracking the body's mass distribution. The distinction between these variables is explicit in research on visual feedback during weight shifting. PLOS ONE: CoP feedback study.
Read the subscripts before calculating. In the first expression, the weights are masses. In the second, they are vertical support forces. Both results have units of metres, so a units check alone cannot catch substitution of one quantity for the other.
Worked example: two supports and one resultant
Imagine a rigid apparatus resting on two narrow supports. Take the left support as x = 0. The apparatus is stationary, gravity is uniform, and the only external forces are its downward weight and the two upward reactions. There are no other applied forces or couples. Treat all forces as vertical and the analysis as two-dimensional.
| Quantity | Left support | Right support |
|---|---|---|
| Horizontal position | 0 m | 0.60 m |
| Upward reaction | 240 N | 360 N |
The total upward force is 600 N. The equivalent contact resultant must preserve both that total and its turning effect:
xCoP = [(240 × 0) + (360 × 0.60)] / 600 = 0.36 m.
The midpoint between the supports is 0.30 m. It is not the correct answer because the forces are unequal. More of the support force acts on the right, so the force-weighted location lies right of the midpoint.
The result does not mean there is a third physical support at 0.36 m. It is a location at which a single equivalent upward force would reproduce the original vertical force and moment. A resultant between separated contact regions can lie where there is no local contact.
Now infer the apparatus's horizontal CoM coordinate. Static equilibrium requires zero net external force and zero net external torque. Its weight is therefore 600 N; taking moments about the left support gives 600 × xCoM = 360 × 0.60. Consequently, xCoM = 0.36 m. The equilibrium conditions are standard mechanics; the apparatus and calculations here are original. OpenStax: static equilibrium.
The two horizontal coordinates match under these assumptions. The calculation has not located the CoM's height. It has not established that mass and pressure are the same physical quantity. It has linked their horizontal coordinates through a complete external-force model.
Change one assumption: add an external support
Keep the apparatus stationary, with weight 600 N and xCoM = 0.36 m. Now add an overhead cable supplying a vertical upward force of 120 N at x = 0.60 m. The original ground supports remain in place. This is a different force system; the earlier reaction readings cannot simply be retained.
Vertical equilibrium gives a new combined ground reaction of 480 N. Moment equilibrium about the left support requires:
Ground moment + (120 × 0.60) = 600 × 0.36.
The ground moment is therefore 144 N·m, and the ground-only CoP is 144/480 = 0.30 m. The corresponding ground reactions are 240 N on each side. Check both sums: 240 + 240 + 120 = 600 N, and (240 × 0.60) + (120 × 0.60) = 216 N·m.
The CoM has not moved. The apparatus has not started accelerating. Yet its ground-only CoP has changed from 0.36 m to 0.30 m. The missing link in an incorrect interpretation would be the additional external support.
This example explains why the measurement boundary matters. A foot-contact record and a record incorporating additional hand support do not describe identical external-force systems. An inference based on feet alone must account for any other contacts before interpreting the relation to whole-body mass. The example does not assume that real hand-support forces are always vertical.
Why a static snapshot cannot answer a dynamic question
A separate issue arises when the system moves. Knowing a position does not tell you its velocity. A mass at a given location can be stationary or moving in either direction. The same position therefore need not imply the same subsequent motion.
Hof's model of walking explicitly combines CoM position with velocity in an extrapolated CoM quantity. Its published abstract supports this narrow lesson about dynamic modelling; it does not establish an individual clinical safety threshold, and the simplified inverted-pendulum assumptions should not be treated as a complete description of every movement. Hof, 2008: abstract.
Avoid turning a force-platform dot into a conclusion about fall risk. First establish whether the record is CoP, a measured or estimated CoM, or another derived variable. Then inspect the task, contacts, coordinate system and time information. A descriptive mechanical calculation and a validated clinical interpretation are separate steps.
Transfer exercise: what does the result actually establish?
A second stationary apparatus has supports at x = 0 and x = 0.50 m. Upward reactions are 180 N and 420 N. Its only other external force is its weight in uniform gravity; there are no applied couples.
- Calculate the ground CoP.
- Decide which CoM coordinate follows from the assumptions.
- A colleague says the result also gives the CoM's height and proves the apparatus would recover after a push. Which parts of that statement are supported?
Reasoned feedback: the force total is 600 N and the moment is 420 × 0.50 = 210 N·m. CoP is therefore 0.35 m from the left support. Under the stated static model, the horizontal CoM coordinate is also 0.35 m. Neither height nor recovery after a disturbance follows. The moment equation contains no CoM height, and the scenario supplies no post-disturbance velocity or response model.
If the right-hand reading were absent, the same calculation would be underdetermined unless another valid constraint supplied the missing information. If the system were accelerating or had another contact, the static inference would need revision. These are changes to the reasoning, not merely changes to the arithmetic.
A practice sentence worth getting right
Try this description: “The contact-force resultant is located 0.35 m from the left reference point. Its equality with the horizontal CoM coordinate follows from the stated static, gravity-and-ground-support model.”
That sentence identifies the measured quantity, the coordinate and the inference. When reviewing your next balance diagram, write those three elements before attaching a clinical meaning to the picture.
Sources
- OpenStax, University Physics Volume 1: centre of mass. Relevant definition and coordinate equations inspected.
- OpenStax: conditions for static equilibrium. Force and torque conditions inspected.
- Effects of Filtering the Center of Pressure Feedback Provided in Visually Guided Mediolateral Weight Shifting, PLOS ONE, 2016. Relevant full-text introduction inspected for the distinction between contact-force and mass-related variables; intervention results are not used here.
- Hof, The “extrapolated center of mass” concept suggests a simple control of balance in walking, 2008. PubMed abstract inspected; use is limited to the model's inclusion of position and velocity.
Sources checked 2 November 2026. All apparatus examples and exercises are original educational constructions.