# Formula and data for EV hill climbing?



## jpheym (Mar 23, 2014)

I'm a new member here. I'd like to rephrase the question from http://www.diyelectriccar.com/forums/showthread.php/happen-electric-golf-car-go-inclineii-93851.html as a data modeling problem.

I'm considering an EV conversion, purchase or DIY. My biggest concern is I live on a mountain. My daily commute first involves driving 5 miles at 35 mph down 1000 feet of elevation, then 6 miles at 35 to 60 mph on typical flat suburban / highway driving. The return trip worries me because I have 1000 feet of elevation to climb at the very end. Can a given EV do it? How much energy in KWH will I need for various speeds? 

So I'd like to numerically model the entire trip, at least to a first approximation. This requires a formula for power consumption.

I came across this post http://www.diyelectriccar.com/forums/showpost.php?p=197619&postcount=23 in which one EV owner said that the formula speed(mph)^2 / 24 = current(amps) fit his vehicle fairly well. Speed^2 is the standard drag equation. Voltage is assumed constant. The constant factor would capture the efficiency of the design (aerodynamics, drivetrain, tires, etc).

So now add in positive elevation change. My guess is the equation for instantaneous power consumed would now be just

power = A*(road speed)^2 + B*(vertical speed).

Has anyone fit an equation like this and determined the constant factors A and B and any residual errors (flaws in the model)? With only two unknowns one should only need a few data points.

Has anyone driven their EV up a known elevation at a known speed and measured total KWH consumed?

Or is this as simple as the following:

potential energy formula: energy (joules) = mass(kg) * height (meters) * g (9.8 m/s^2) 
or:
energy (KWH) = mass(kg) * height (meters) * g (9.8 m/s^2) * 2.77e-7 kwh/joule
or:
power (KW) = 1e-3 * mass(kg) * height(meters) * g / time(seconds).

So for a 3300 lb EV to climb 1000 feet:
1497 kg * 305 meters * 9.8 m/s^2 * 2.77e-7 kwh/joule = 1.22 KWH

... which seems not so bad since EVs typically get 0.5 to 3 mile/KWH (flat ground). So 1.22 KWH extra to climb 1000 feet is the equivalent energy of driving 0.6 to 3.6 extra miles. This seems reasonable. Am I on the right track here?


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## sergiu tofanel (Jan 13, 2014)

The extra energy required to climb a certain height is given by E = m*g*h. 

In metric, the formula yields (1500kg)*(9.8m*s^-2)*(300m), which is 4.4MJ, or 1.2KW*hr.

However, the issue is not battery capacity, but rather motor power. If you plan to climb that hill in let's say, 2 minutes, you need 4.4MJ/120s, or approx 50HP extra!!!!!


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