A bike and a rider match well when the centre of gravity of the rider, is above the pedals/the bracket/etc.

I have noticed some simple math gives a much more accurate estimate of the point of gravity, than a visual estimate.

My simple math:

Head = 9% of total body weight

One arm = 9% ( upper arm = 50% )

Upper body = 36%

One leg = 18% ( upper leg = 66 % )

In the Fit Advisor model all body parts have simple pivot points.

In the Fit Advisor model the pivot point of the upper body is right above the setback point of the seat post plus adaptation for sitting position.

That makes the calculation manageable.

The centre of gravity of the upper body is 0.5 * 0.36 * bodyweight * ( B - C ) * sine (crouch angle) away from the seatpost SB

B and C are variables of the Fit Advisor model

I have made an Excel spreadsheet to calculate one centre of gravity for head, body, legs, arms.

My question: would it be possible to pinpoint an estimated centre of gravity in the Fit Advisor model?

Thank you.

I'm open to doing something like this. However, I don't fully understand your sample calculation. Normally, a calculation for center of gravity would sum the contributions from individual elements of mass but all of those contributions would be divided by the total mass, so your final result would not include any units of mass. Your sample appears to calculate a number with the units m kg. You've also inserted a sine function in the equation. I assume this was to calculate the Y component for center of gravity while perhaps the you'd use a cosine function to determine the X component.