I used to perform a lot of physics calcs in my youth, but not of these types; itÃ¢â‚¬â„¢s all new to me. Thanks for spot-checking meÃ¢â‚¬Â¦ IÃ¢â‚¬â„¢ve been at this for days with spreadsheets and no one to talk to.
OK, I edited/corrected the formula notation above. On we goÃ¢â‚¬Â¦Calculate Steady State:
, solve for V
V = P / I -> 1491.4 / 44 = 33.9
This is amusing. From the relationship of P
we can play around and manipulate the factors until we arrive at suitable products for the system.
The part that bugs me is that the Motor Torque Constant
) is relative and not constant at all unless we choose it to be so for control. In other words, if I am going 30 mph (er, sorry 48.3 kph
) for hours on level ground and my battery drops voltage then Kt
must change. Similarly, if I have a current limit set and along my journey I encounter wind or an incline, Kt
must change. Perhaps itÃ¢â‚¬â„¢s a moot point but it has had me twisted up Ã¢â‚¬â€œ at least until I could explain it: K
is an ideal theoretical value only. Do you agree? Meat & Potatoes:
The whole point of this exercise is to mathematically anticipate the requirements for a steady-state wheel in motion given three values: diameter
, and power
consumed. I think weÃ¢â‚¬â„¢ve flogged the observed system enough, therefore I wish to analyze the internal workings and construct a model of the electromotive forces to complete the energy balance equation.A single straight wire
passing through a uniform magnetic field produces a force, (formula derived from the Lorentz
force equation F
F = IL x B, where
F = Force in Newtons (N)
I = Current in Amps (A)
L = Length of wire in Meters (m)
B = Flux density in Telsa (T) or Gauss (g x 10,000)
Given as Torque:In a single electric circuit
loop, turn, or winding), we have two sides pulling in opposite directions along a common axis, therefore:
For the sake of discussion let us assume that r
= Â½ d
of the original wheel. The value for Flux density (B
) is picked arbitrarily (we can discuss calcs a bit later).
r = 12 inches / 0.3048 meters
B = 0.5 T
I = 44 A
Ãâ€ž = 33.9 Nm
Solve for L
L = Ãâ€ž / 2rBI -> 33.9 / (2 * 0.3048 * 0.5 * 44) = 2.53 mObservations:
- This is a single-phase solution.
- The frequency would be 420.2 rpm or 7 Hz.
- The conductor would have to be 11 AWG to carry 44 A safely.
The heat generated by the system would be as follows:
R = 4.1328 ohms/km for 11 AWG -> (4.1328 * 2.53) / 1000 = 0.01046 ohms
R = V / I
P = I^2 * R = V^2 / R
P = 44^2 * 0.01046 = 20.2 watts
Calculate the Efficiency:
Pe = (Pi Ã¢â‚¬â€œ Pr) / Pi -> (1491.4 Ã¢â‚¬â€œ 20.2) / 1491.4 = 98.6%, where
Pe = Power, Efficiency
Pi = Power, inital or imperical
Pr = Power, resistance
This is probably the best this theoretical system will ever see. To overcome the loss of efficiency the system would need to provide > 1.4% more power.Summation:
In the first post of the thread I calculated the Angular Velocity (Ãâ€°
) and Torque (Ãâ€ž
) given wheel diameter (d
), velocity (v
), and power (P
) used. The second half of this thread develops a model which predicts the electromotive forces for a single circuit. Thus we can represent the elementary Energy Balance Equation
Efinal = Ein - Eout, where**Calculation:
Ein = (1512w @ 44A x 34.4V) Ã¢â‚¬â€œ (20.5w required to overcome electrical resistance**); Work on the system, constant current
Eout = 1491.4w; Work by the system
Efinal = 0; balanced
1491.4w / 98.6% = 1512w; 1512 - 1491.4 = 20.5. If current (I
) is constant then V
must rise to 34.4V.
EDIT: Corrected EBE
Are you still with me? Are my calculations correct?
Thanks for checking, KF