Fletcher wrote: (I believe Gearing is squared in the formula because KE Output for CoE is a squared term).
e.g. for Driver mass 4.0kg and Lateral Load mass of 0.5kg with 4 x Gearing ...
* Driver Acc = 'g' / ( Gearing^2 * Lateral Load mass to Driver mass Ratio + 1 ) => -10 / ( 4^2 x 0.125 + 1 ) = -3.333 m/s^2
The Lateral Load Acc is 4 x the Driver Acc because of the gearing used.
Fletcher wrote: Can someone please explain to me why we have to square the Gearing factor when considering the inertia quotient effect on the new acceleration of the Driver ?????????
Fletcher,Fletcher wrote:I hope you will both have a say in explaining to me the inertial factoring of my formula's.
Your driver acceleration equation is identical to the Atwoods acceleration equation.
The derivation of the Atwoods acceleration equation for 3 hanging masses was done in this post here using free body diagram analysis:
http://www.besslerwheel.com/forum/download.php?id=10709
If you set m2 and m3 to zero, you get a simpler form of the equation as shown in this diagram here:
http://www.besslerwheel.com/forum/downl ... er=user_id
The acceleration equation for one hanging mass is:
a = (g * driver_mass) / (driver_mass + I/r1²)
This equation is basically: a=F/m
Linear acceleration = Linear Force / Linear Mass
- The acceleration is the linear acceleration of the driver mass.
The linear force on the system is the driver mass x acceleration due to gravity.
The linear mass of the system is: (the driver mass) plus (the moment of inertia of the Flywheel divided by the radius squared), where the radius is the radius of the driver mass.
Note: when you take the moment of inertia of the flywheel and divide it by the driver radius squared, you get the "linear mass" of the flywheel that is felt by the driver mass during the acceleration.
To equate this to your gearing problem, the flywheel rim mass represents your lateral load mass.
If you take the right hand side of the equation and divide the numerator and the denominator by the driver_mass, you get:
a = g / {1 + [I/(driver_mass * driver_radius²)]}
Your lateral load mass is the flywheel rim mass. It has a "rotational mass" or moment of inertia of "I".
If you take a moment of inertia and divide it by a radius squared, you get a linear mass.
So the denominator of this acceleration equation is looking a lot like:
(lateral_load_mass/driver_mass) + 1)
But we need to do a little more algebra to get the gearing factor in there.
r1 = radius of driver mass
r2 = radius of flywheel rim mass (or lateral load mass in your case).
If you take r1 and multiply it by some scaling factor (or gearing factor), you get r2.
r2 = C * r1
or:
r2² = C² * r1²
almost there ...
The moment of inertia of a rim mass is I = mr²
If your load_mass was a point mass at radius r2, the moment of inertia of your load_mass would be
I = load_mass * (r2)²
or
I = load_mass * C² * r1²
so the denominator term of the acceleration equation would be
{1 + [I/(driver_mass * driver_radius²)]}
={1 + [load_mass * C² * r1²/(driver_mass * r1²)]}
={1 + [load_mass * C² /(driver_mass )]}
={1 + [C² * load_mass /(driver_mass )]}
Which would make my acceleration equation equal to:
a = g / {1 + [C² * load_mass /(driver_mass )]}
Which matches your acceleration equation of:
Driver Acc = 'g' / ( Gearing^2 * Lateral Load mass to Driver mass Ratio + 1 )