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 you'd use a cosine function to determine the X component.
Thank you for your comment.
You are quite right.
My calculation in Excel is more elaborate.
In my request I did not want to put apart the complete calculation
I just wanted to say elementary math will do for me.
Not like the dynamic equations with (in)stable eigenvalues by JDG Kooijman in Science.
http://science.sciencemag.org/content/332/6027/339
I just want to check the common 55/45 rule (55% of the weight should be on the rear wheel)
Sine and Cosine
In my Excel spreadsheet I use the Torso Angle value from the Dimensions/Rider dialogue box, which is 0 dgr for a flat riding rider.
Then the crouch slider in the Rider dialogue box is to the far right, which conveys the impression on me that Torso Angle is a deviation from the upright sitting position
The choice for sine or cosine is determined by the defnition of the crouch angle.
I prefer the Torso angle value from the Dimension/Rider dialogue box.
I have measured the knee angle for the left leg with right click custom dimensions in the display.
I have not measured the hip angle for the left leg, but I have measured the angle between left upper leg and right upperleg in the display. Then it will be less elaborate to calculate a new center of gravity after a change of the crouch angle.
I am sorry for the confusion I may have raised.