Putting the PHV through it's paces in the hot desert... (Rick's turn for the PHV)

Interesting, it looked really difficult to stay in ev mode accelerating to 60. It would be a nice improvement if they had an ev button that could lock it in that mode, pressing the accelerator hard could operate force a change mode like it does to kick down in an automatic transmission. If the ECU allowed multiple of the packs to power the car it also seems that the phv would safely increase its acceleration.

How did you calculate time to 60? Do you have the losses for drag, friction, and electricity?

All I could come up with was

Energy = Kinetic Energy + ME(mechanical losses durring accelleration) + electrical losses durring accelleration.

Pt = 1/2 MV^2 + MF(V^2,t) integrated over time + EL (P,t) integrated over time.

assuming that power is constant. But I don't know how much the mechanical or electrical losses are.

You should be able to pull 200 A through 6 awg wire safely for a short period of time. Toyota specked the GIII to provide 27 kw at peak so the power is higher. I would also assume that the 60kw spec on the mg2 is also engineered with a safety margine. I do not know what the peak power through the inverter is though. In EV mode what is MG1 doing power wise? Just geeking out. My guestimate is the current hardware with the addition of the standard ev button and software changes should be able to hit 60mph in less than 20 seconds.

 

Can you get MG1 current? I still would like to confirm if MG1 is assisting (consumpting electricity) during acceleration (in negative RPM).

 

Is my calc so far off no one is commenting out of politeness ?

 

AG, I calculated this way (with staring values):
Power(p) in watts
Force(f) in Newtons
Acceleration(a) = 1 meter/sec*sec
Mass(m) = 1450 kg
Time(t) = 30 seconds
Distance(d) in meters
Work(w) in Newton*meters

f=ma
w=f*d, substituing d = at^2/2, we have w=fat^2/2
p=w/t


Substituting fat^2/2 for w and cancelling a 't' gives p = fat/2.
Our case conveniently had a=1, so p in watts simplifies to mt/2

 

acceleration is not a constant, it decreases with velocity if power is held constant, so your formula for distance is wrong. Your formula also doesn't account for any losses.

 

AG,
You may be right, I thought I could use average acceleration. The power calculated would be net at the wheels.

 

(P) Power = (F)Force * (v)velocity
P = Mass(M) * acceleration(a) * v
a=dv/dt

P = Mv * dv/dt

which is quite solvable to if P is a constant and we integrate between velocity of 0 and V

PT = MV^2/2

This formula also says the change in kinetic energy of the system is equal to power * time applied, which is the energy added to the system. Mass of the car is also a constant since we are much slower than the speed of light. The power must also overcome drag, friction, and electrical losses. I'm sure someone out there knows something about these figures. Power is really a function of motor speed, so the power curve of the motor when taken into account would increase this time. Given this motor though making a Power a constant is a good estimation

 

Its a direct formula to calculate power from torque and rpm. There also is a constant formula between rpm and vehicle speed. So the only missing information is voltage and you can solve for current. If you are concerned with power at the battery you can simply use voltage of the battery pack. If you like I can look up the constant for ft-lb*rpm to watts.

 

I definitely don't want to make you violate anything to satisfy my curiosity.

I'd like to see mg1 and mg2 torque at 50mph and 60mph on flat roads. You've provided some really good information.

 

The first video told me a lot. I'd also like to see the torque at constant speed. That would give me a better idea of kwh/m at that speed. Which ever 2 speeds you normally want to cruise on electric. I wanted to get a high speed and a lower speed to see how the losses affect the car.

 

Yes, this is the approach I used. Plugging in numbers for
PT = MV^2/2
gives the same result I found of 21.7 kw.

I plugged in the end numbers of 30 meters/sec and 30 seconds. We could plug in intermediate values say at 10 and 20 seconds to verify the constant power assumption.

 

Yes, that would settle my theory. If the ICE is off and you see torque in MG1, that would suggest motive force to the wheels.

 

sagebrush,
I hadn't fully thought about the averaging. You should get the correct numbers for average power no matter what the acceleration is, since the same work is performed to get the car up to speed other than losses no matter how it accelerates.

Conceptually it just doesn't look right. Just remember not to do the common mistake of thinking twice the power will accelerate a car in half the time. It takes four times the power.

 

I promise

 

If DC/AC conversion and gearset losses add up to 10%, and power to roll the tires averages say 3 kw, the battery would be discharging about 28 kw.

So says my detailed calcs, anyway ;-)

As an aside, I know people keep hoping to shoehorn the Prius PHV into a no-matter-what EV for 14 miles, but that is just not the design objective, and I'm glad it is not. The cost would be prohibitive.