Yes, I agree they will each increase at different rates but they will each be with the square of RPM.
Coil1 = K1*RPM^2 Coil2 = K2*RPM^2
Coil1+coil2 = (K1+K2)*RPM^2 = K*RPM^2
So in the end the power curve is with the square of the RPM. This means it can be reproduced by another dual rotor machine of standard design.
"A little elaboration on matching curves may help. The general curve for a cubic is: P = AxV^3 + BxV^2 + CxV + D. In this D = 0."
Yes this is true - the power in the wind follows the cube of wind velocity, but B, V, and D are 0.
As far as the power curve of the blades go we really don't know the formula for it - it could even be a fourth power, it really depends on the blades, - one thing we can say is that it is concave down everywhere.
In matching the alternator to the blades the 2 curves that have to intersect are : 1) the alternator power verses RPM and 2) blade power at the shaft verses RPM.
The blade power curve is concave down everywhere and the alternator power curve is K*PRM^2. They can only intersect at 2 places - one of them being (0,0).The sooner you start the longer it takes.[ Parent ]
When the alternator hits 150 rpm the second highest pulse becomes large enough to cause the diodes to conduct. That speed is the first break point. So now we have a second pulse causing conduction. Its value is 2/3 that of Vp. So, for that pulse we may write Ib2 = (2/3xVp - vb)/Ra. The energy delivered by that pulse is given by Eb2 = PtxVbx(2/3xVp - Vb)/Ra. The energy from the alternator during that pulse is Ea2 = PtxVpx(2/3xVp - Vb)Ra. Since both pulses now contribute, we must add their contributions to get the total energy into the battery, which gives:
Eb = PtxVbx(5/3xVp - 2Vb)/Ra. Ea = PtxVpx(5/3 - 2Vb)/Ra.
This applies from 150 rpm to 300 rpm.
Finally, when the alternator's speed hits 300 rpm, the second break point, the third pulse will be large enough to cause the diodes to conduct. This pulse has an amplitude of 1/3 that of Vp. We now get a current of Ib3 = (1/3xVp - Vb)/Ra, for an energy value of Eb3 = PtxVbx(1/3xVp - Vb)/Ra. Ea3 = PtxVpx(1/3xVp - Vb)/Ra. The other two pulses are also contributing so we need to add this to the contributions from the others. We get: Eb = PtxVbx(2xVp - 3xVb)/Ra. Ea = PtxVpx(2xVp - 3xVb)/Ra. You get the average power by adding up the energy contributions from all the pulses per cycle (there are two of each) and dividing by the time in seconds per cycle. These equations show that the rate at which the alternator puts power into the battery as a function of rpm is increased by 2/3 after the first break point and is twice as great after the second break point as its initial value. This is the basis for the power curve that I gave. GeoM[ Parent ]
I'm not sure why I didn't get it before, probably because I'm not used to thinking about batteries - strictly heating for me so far.
It is not easy to analyze, but interesting to think about.The sooner you start the longer it takes.[ Parent ]
