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Okay, here's the specs (approximate) and calculations I've done

-Total mass (after passengers and everything) = 1000kg (W = 9800 N)
-Coefficient of drag = 0.4 source
-Frontal area = 2 m2
-Maximum required speed = 60 kmph (16.6 mps)
-Maximum expected angle of hill = 15 degrees
-Maximum required speed at 15 degrees = 20 kmph (5.5 mps)
-Coefficient of rolling resistance = 0.0100 source
-Density of air ρ = 1.125 kg/m3
So, ∇z (vertical velocity at 15 degrees) = 5.5 Sin(15) = 1.42 m/s

Power required=M·ν·a + 0.5·ρ·CD·A·ν3+M·g·∇z+μ·M·g·ν
At steady state a = 0m/s2

On flat ground at 60kmph
Power = 3929 W = 3.93 kW

On incline (20 kmph and 15 degrees)
Power = 14801 W = 14.8 kW

My problem is, are you supposed to get such a massive discrepancy between those two numbers? I've driven heavier and underpowered cars on similar inclines (I live in a mountainous country) before and I can go faster than 20kmph with less than about 18 horsepower.

My end goal is to convert a Maruti 800 into an electric car and I want to use the least power consuming motor possible, mainly for budget reasons.

ywa1n
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  • Check your units and numbers - density has a problem. – Solar Mike Mar 25 '22 at 05:44
  • My car uses a lot more fuel climbing mountains than driving on flat countryside... – Solar Mike Mar 25 '22 at 08:06
  • Yeah, the density units has a problem, fixed. It would take more energy obviously but I'm more interested in the power of the motor needed. – ywa1n Mar 25 '22 at 10:06
  • So, without checking your numbers, the motor power has to be at least 14.8 kW. Have you accounted for drivetrain losses? – Solar Mike Mar 25 '22 at 10:25
  • I have not, but I wouldn't mind losing some acceleration, top speed or hill climb performance if it saves me from measuring drivetrain efficiencies. That stuff can be stressful – ywa1n Mar 25 '22 at 10:52
  • Does this help? https://engineering.stackexchange.com/questions/31501/how-can-i-calculate-the-power-and-torque-required-for-the-motor-on-a-wheeled-rob/31502#31502 Air drag was not considered here but you just add that to the force of rolling friction Froll. – DKNguyen Mar 25 '22 at 14:54
  • And +1 for actually having most of the relevant specs worked out which is not typical for this kind of question. – DKNguyen Mar 25 '22 at 14:59
  • "*I've driven heavier and underpowered cars on similar inclines (I live in a mountainous country) before and I can go faster than 20kmph with less than about 18 horsepower.*" If your engine can only produce enough torque to maintain speed up the hill, the speed you go up is the speed you hit the hill at after accelerating on level ground. Because objects in motion stay in motion if you remove losses (i.e. just enough engine power to cancel weight and friction.) Actually accelerating up the hill is different. – DKNguyen Mar 25 '22 at 15:21
  • Did you drive *electric* cars like that? Or did you drive normal internal combustion cars, with a normal gearbox, allowing to convert quite meager power into quite a bit of torque. – SF. Mar 25 '22 at 15:28
  • Yeah I drove an IC car up the hill, not an electric one. I'm assuming you don't have to accelerate too much uphill from speed. – ywa1n Mar 28 '22 at 04:18

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Your 20 degree hill would place it in the running as the steepest street in America. I'm surprised the difference isn't greater.

Tiger Guy
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    This seems to be more of a comment than an answer. – Eric S Mar 25 '22 at 16:01
  • Yeah this is not an answer, should be a comment. That being said, where I live, 20 degree slopes aren't common but they definitely are not rare either. – ywa1n Mar 28 '22 at 04:20