Range Estimation:

 
range[km]=250 x capacity[kWh] / (mass[kg]^0.6) 
-or-
range[miles]=250 x capacity[kWh] / (mass[lbs]^0.6)

Wire Temperature? :

Thot = Tcold + (K x ((Rhot - Rcold) / Rcold)) 
 
where K = 256.4 for copper 
       Thot = hot temperature in deg.C 
       Tcold = cold temperature in deg.C 
       Rhot = resistance at hot temperature in ohms 
       Rcold = resistance at cold temperature in ohms

Horsepower:

Hp=(torque*RPM)/5252

AWG to Metric Conversion

Current Limiting Formula for LEDs

Amps, Volts, Watts and Watt-Hours 101

[sl.whiteside]

Is the torque in ft-lb, in-lb, or n-m?

[rbgrn]

Originally Posted by sl.whiteside

Is the torque in ft-lb, in-lb, or n-m?

Ft-Lb I believe.

[sl.whiteside]

Originally Posted by rbgrn

Ft-Lb I believe.

Robert,
I confirmed after checking my Pocket Ref by Thomas J. Glover, 2nd Edition.
It is Ft-LB.

[rbgrn]

I love that book, btw.

[Thalass]

In the range equasion (range[km]=250 x capacity[kWh] / (mass[kg]^0.6) ), what is the "0.6"? It looks like me to be CoD, but there's no real reason for that, just a hunch.

[weelliott]

I am not familiar with the formulas, but I can tell you that it is not from Coefficient of drag. The fact that the exponent is under 1 means that the mass has a relation that factors it in less than what it would factor in as if you were looking purely at acceleration--where weight matters.

This equation is very simplified. If you were to develop an equation for range that assumed that the car were traveling on a flat surface at a constant speed, weight would not matter at all, but drag area would drive the results.

Conversely, if you were to base range on a stop and go city type environment where it is mostly acceleration and aerodynamic losses are relatively minimal compared to acceleration and Rolling Resistance, then the weight of the car would matter more, and that exponent would be 1. so it looks like some assumptions about an average driving cycle have been made and the 0.6 exponent has been used to reflect that compromise. It looks as though it uses some stop and go or some hills.

The fact that Cd is not in the equation suggests to me that this assumed driving cycle doesn't have much traveling over 45 mph.

So I am really curious. What is the assumed cycle? Is there an assume value for drag area? (Drag area is just Cd times frontal area, the two variables specific to a car's design.)

[Vwbeamer]

^ is powers. You need a scientific calculator, it will have a ^ key.

BTW the formula is very optimistic. for example a 2500 lb car with a 10KW power pack would go 91 miles according to the formula.

[Franky.EV]

I've try to put all formulas in a single oooCalc spreadsheet.

Look at :

http://www.diyelectriccar.com/forums/showthread.php/yes-another-ev-calculator-45278.html

It takes into account : grade, DOD, motor torque, gear, shift time.

[Crash]

I don't think that range formula is all that accurate...

I was trying to figure out the range a 3000Lbs Corvette with a 40.2kWh pack would go and it's saying something around 82mi... That doesn't sound right at all unless I'm doing my math wrong:

range = (250 * 40.2) / (3000 ^ 0.6) = 82.394mi

That really doesn't sound right.

Tesla Roadster is 2723Lbs with a 53kWh battery pack (I know this has changed lately, but lets go along with the old numbers for now.)

Using the same formula I get 115.130 which is 135miles less than what Tesla claims.

I think that 0.6 is the wrong number. It should probably be closer to 0.5. 0.5 would actually put the range for the Tesla closer to 250mi.

Running the numbers with the 3000Lbs Corvette and a 40.2kWh pack comes out to 183mi. That makes more sense to me. Maybe still a little high of an estimate, but sounds more accurate than 82 miles out of a 40kWh battery pack.

EDIT: I forgot to mention, the original formula is suggesting that my car would use 490wh/mi.