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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.

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  • $\begingroup$ Check your units and numbers - density has a problem. $\endgroup$
    – Solar Mike
    Mar 25, 2022 at 5:44
  • $\begingroup$ My car uses a lot more fuel climbing mountains than driving on flat countryside... $\endgroup$
    – Solar Mike
    Mar 25, 2022 at 8:06
  • $\begingroup$ 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. $\endgroup$
    – ywa1n
    Mar 25, 2022 at 10:06
  • $\begingroup$ So, without checking your numbers, the motor power has to be at least 14.8 kW. Have you accounted for drivetrain losses? $\endgroup$
    – Solar Mike
    Mar 25, 2022 at 10:25
  • $\begingroup$ 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 $\endgroup$
    – ywa1n
    Mar 25, 2022 at 10:52

2 Answers 2

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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.

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    $\begingroup$ This seems to be more of a comment than an answer. $\endgroup$
    – Eric S
    Mar 25, 2022 at 16:01
  • $\begingroup$ 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. $\endgroup$
    – ywa1n
    Mar 28, 2022 at 4:20
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Firstly better calculate rotational speeds and torque required to move your car, not power. And if you need power, you can calculate it from torque and speed.

  1. Transform linear car speed you need to reach into the rotational speed of its wheels. Next if there's mechanical parts like gears or something you should calculate rotational speed on your motor. If there's several gears (there's not only reducer but also you can use gearbox), calculate speed on higher gear. Also it's better to know highest speed on all gears.

  2. IDK about drag coefficient (since its low influence on calculations) but you should calculate required torque on your motor. Sum up all forces interfering moving (gravity influence on 15 degrees uphill, inertia etc) and transform it into the torque on your wheels. After that, just like with speed calculations, if there's several gears calculate torque on each. Most important here is lower gear.

Its oversimplified, but you can get the point. After obtaining torque and speed you should choose the motor. Often it chooses not by power, but by torque and speed. It seems like choosing by power is the same as choosing by torque and speed, especially knowing that rotational speed*torque = power of motor, but it is wrong. You should "cover" torque and speed. Choose motor which can work with required torque on lower gear and speed on higher gear. If your conversion is about fixed gear (w/o gearbox), you should choose motor which work with required torque and speed at the same time, but it is not rational.

To choose motor properly you should obtain torque/speed curves (mechanical graph) required to reach needed parameters. You can do this the same way as I already described (calculate speed and torque in several points: from 0 speed and highest torque you need to start moving in hardest required conditions to highest speed with lowest torque required to keep in on the highest speed).

There's torque/speed/power curves for motors. You should choose motor which has "higher" curve, than your mechanism has (area under your "required" graph has to be less than area under motor graph). If you do so, it will mean that your motor has power to provide required characteristics. Note that motors has several "zones" of control: nominal and flux weakening also known as power output zones which can allow you to obtain more rotational speed when you don't need any considerable torque.

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