How can I accurately determine the necessary wattage for a motor in my vehicle?

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Discussion Overview

The discussion revolves around determining the necessary wattage for a motor in a vehicle, focusing on calculations related to vehicle mass, load, rolling resistance, and power requirements. Participants explore various formulas and parameters affecting motor performance, including coefficients of friction and drag, while addressing practical experiences with motor usage in electric vehicles.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant presents vehicle specifications and initial calculations suggesting a required motor wattage of 4KW, yet notes successful operation with two 0.5KW motors.
  • Another participant questions the validity of the calculations, particularly regarding the coefficients of rolling friction and drag, suggesting they may be too high.
  • Concerns are raised about the formula used for calculating forces on a slope, specifically the inclusion of the cosine function and its implications at steep angles.
  • Participants discuss the actual power consumption observed in practice, with one noting a consumption of 24V and 24A, leading to confusion about how this relates to calculated power requirements.
  • There is a mention of the motors' RPM and gear ratios, with one participant providing specific RPM values and expressing curiosity about the implications for power calculations.
  • Some participants express uncertainty about the correct coefficients to use in calculations, indicating a need for further clarification and validation of the values used.
  • One participant suggests an alternative approach to the force calculation on an incline, proposing a different equation structure that may better represent the forces involved.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the appropriate coefficients for rolling resistance and drag, nor on the validity of the force calculation methods. Multiple competing views remain regarding the interpretation of the motor's performance and the calculations presented.

Contextual Notes

There are limitations in the assumptions made regarding coefficients of friction and the forces acting on the vehicle, as well as unresolved mathematical steps in the calculations presented. The discussion reflects a range of experiences and interpretations of the physics involved.

Who May Find This Useful

This discussion may be useful for individuals interested in electric vehicle design, motor performance analysis, and those seeking to understand the complexities of calculating power requirements in vehicular applications.

Gnana guru
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Hi fellas,
Really how to select the wattage of the motor as I'm stuck with that. I came across several formulae, yet still unable to do it so.

Vehicle Mass = 120Kg
Load Mass = 100Kg (max)
Front tyre = 8" x 2
Back Tyre = 12" x 2
Tyre material = Polyurethane
Dimension = 25" x 35"
Speed = 9.3km/hr

They told that,
coeffient of rolling friction for PU = 0.2-0.7
coeffient of drag = 0.04 ( for oval shaped bodies)
density in aerodynamics force = density of air = 1.225

By following attached calculations I've got 4KW, yet with two - 0.5KW DC brushed motors itself it is working. I just want to know how it is possible!

https://drive.google.com/open?id=1dTDyLQQYv1AD3SHYttNombKaIr7vTttf
 
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Gnana guru said:
By following attached calculations I've got 4KW, yet with two - 0.5KW DC brushed motors itself it is working. I just want to know how it is possible!
A motor can do more work than rated for a short time, but it will overheat and fail later.
How do you know it works?
The greatest power component is expected to be when it is climbing a gradient as maximum speed.
Will it climb a gradient at full speed? What current, voltage and power does it then require?
 
Baluncore said:
A motor can do more work than rated for a short time, but it will overheat and fail later.
How do you know it works?
The greatest power component is expected to be when it is climbing a gradient as maximum speed.
Will it climb a gradient at full speed? What current, voltage and power does it then require?

Thanks for the reply !

As a matter of fact the motor is working fine for a very long time without any problems. To be frank, I'm a newbie in a company where for their E-vehicle they are using this. All the calculations that I've specified are from their vehicle model only. Yet I'm looking for a justificaition on "How they are using them in their vehicles?"
It can take upto 12degree of slope's without any problem. Ideally for 75Kg load, it consumes nearly 24V,24A.
 
##F_t=F_r + F_g + F_d + F_v##
##F_t = μN + \frac {μN } {cos(θ)} + C_d \frac {ρv^2} {2}A + (ma +Ia) ##

For rolling resistance, your range of μ = 0.2 - 0.7 and you take 0.4. I have no experience of this area, but these figures sound very high to me - as if you're dragging a sledge rather than a wheeled vehicle.. WikiP* quotes 0.07 as a worst case for 19C stagecoach wheel on dirt road with soft snow. Ordinary car tyres on sand is 0.3. (* Maybe wildly guessed? But it shows the order of values expected of wheels by someone with an interest in that area.)

For force due to gradient you use ##\frac {μN } {cos(θ)}## I don't understand why you divide by ##cos(θ)##? If the slope were vertical, that force becomes infinite, rather than just equal to the weight (since you use 1 for cos(θ) , I assume that's horizontal and θ=0.) Also, I don't see why μ comes in here. In fact the normal force decreases as the slope increases, so perhaps you should look at the first two terms together.

## F_d## seems ok, but I can't understand your area. It seems to come to over 20 m2 !

I'm not sure what your last term, Equivalent Inertial Force, is, but since most of your problem comes from the two 789 N terms, I won't bother to look into it yet.
 
On the back of an envelope;
120kg tare, 75 kg load = 195 kg mass.
9.3 km/hr = 2.583 m/s road speed, Vdiag (parallel to slope).
12° slope. Sin( 12° ) = Vup / Vdiag = 0.2079 ∴ Vup = 0.2079 * Vdiag = 0.537 m/s
Force = 9.8 * 195 = 1911. N
Distance vertically per second = 0.537 m
Energy = 1911 * 0.537 = 1026.2 joules per second = 1026 watt.

You claim “It” consumes 24 amp * 24 volt = 576. watt.
So that must be the power to each motor, making the total power = 1152. watt, to do 1025 watts output.
Check, is that the current to each motor?
 
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what is the motors' rpm at the 12" wheels' 164 rpm at 9.3 km/hr?

PS, I realize this has no effect on the power calculation, I am just curious.
 
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Merlin3189 said:
##F_t=F_r + F_g + F_d + F_v##
##F_t = μN + \frac {μN } {cos(θ)} + C_d \frac {ρv^2} {2}A + (ma +Ia) ##

For rolling resistance, your range of μ = 0.2 - 0.7 and you take 0.4. I have no experience of this area, but these figures sound very high to me - as if you're dragging a sledge rather than a wheeled vehicle.. WikiP* quotes 0.07 as a worst case for 19C stagecoach wheel on dirt road with soft snow. Ordinary car tyres on sand is 0.3. (* Maybe wildly guessed? But it shows the order of values expected of wheels by someone with an interest in that area.)

For force due to gradient you use ##\frac {μN } {cos(θ)}## I don't understand why you divide by ##cos(θ)##? If the slope were vertical, that force becomes infinite, rather than just equal to the weight (since you use 1 for cos(θ) , I assume that's horizontal and θ=0.) Also, I don't see why μ comes in here. In fact the normal force decreases as the slope increases, so perhaps you should look at the first two terms together.

## F_d## seems ok, but I can't understand your area. It seems to come to over 20 m2 !

I'm not sure what your last term, Equivalent Inertial Force, is, but since most of your problem comes from the two 789 N terms, I won't bother to look into it yet.
Hi Sir,
1. Yes that might 0.4 might be problematic in calculation as I'm not sure what would be the correct value. I'll look to it anyway. Thanks for pointing out that 0.3 Wiki page

2.Check this link sir...
https://drive.google.com/open?id=1fBPutNWXP2f05GAYpp2ytUeYPUmnJ49D

I'm convinced with this one so that only I went with cos(θ) in denominator.Yes as the inclination goes to 90degrees, its impossible to make a straight 90degree climb by any vehicle

3.Its E-vehicle with length = 25" ,width = 35"

I even tried with 0.2 for coeffiecnt of rolling resistance, yet I ended with 2.7KW. Still it doesn't solves my doubt
 
Baluncore said:
On the back of an envelope;
120kg tare, 75 kg load = 195 kg mass.
9.3 km/hr = 2.583 m/s road speed, Vdiag (parallel to slope).
12° slope. Sin( 12° ) = Vup / Vdiag = 0.2079 ∴ Vup = 0.2079 * Vdiag = 0.537 m/s
Force = 9.8 * 195 = 1911. N
Distance vertically per second = 0.537 m
Energy = 1911 * 0.537 = 1026.2 joules per second = 1026 watt.

You claim “It” consumes 24 amp * 24 volt = 576. watt.
So that must be the power to each motor, making the total power = 1152. watt, to do 1025 watts output.
Check, is that the current to each motor?

Hi ,

This looks quite compromising but what is the real question is "I've measure the Ampere from the battery pack side. i.e., From battery to controller, from controller to separate motors. Hence battery is the only source of power, both the motors tool only 24A that slightly confuses me. Anyway I'll look to it and get back to you about each motor's current consumption. But I thinks mostly both might have consumed only 24A"
 
JBA said:
what is the motors' rpm at the 12" wheels' 164 rpm at 9.3 km/hr?

PS, I realize this has no effect on the power calculation, I am just curious.

Hi fella,

Motors RPM is 5300 and there is a gear ratio of 32:1, which results in 165 RPM. Stay curious buddy!
 
  • #10
Gnana guru said:
I'm convinced with this one so that only I went with cos(θ) in denominator.Yes as the inclination goes to 90degrees, its impossible to make a straight 90degree climb by any vehicle

But close to 90 and the force should be close to the weight where as yours is close to infinite. This suggests something is wrong with the equation.

See also..
http://thecraftycanvas.com/library/...pers/incline-force-calculator-problem-solver/

The equation should be something like

Fr + MgSin(theta) +...

where Fr is the rolling friction force and MgSin(theta) is the force of gravity acting down the slope.

I haven't looked for other issues.
 
  • #11
CWatters said:
and MgSin(theta) is the force of gravity acting down the slope.
Baluncore said:
12° slope. Sin( 12° ) = Vup / Vdiag = 0.2079 ∴ Vup = 0.2079 * Vdiag = 0.537 m/s
That is fixed by the envelope calculation in post #5. The road speed is "on the slope", not horizontal.

Why the measured power input is only half that needed to climb the slope at top speed is not yet explained.
 
  • #12
I'm thinking your diagram is from https://www.plantengineering.com/si...andling/2ee09d470097e2678c18b480e828e1fd.html

Unfortunately I can't find how they derived the rule. The only way it makes any sense at all to me is, if you continue to push horizontally as you move up a slope. This might be a reasonable description of pushing a trolley along a horizontal surface and encountering a bump or short ramp. Then, if you met a vertical surface, obviously the resistance is infinite (or at least indefinite and large.)
If there you pushed parallel to the surface, you'd be pushing vertically, carrying the full weight, and the rolling resistance would be zero.

I suspect they are also including in their "rolling resistance" your contribution to raising the mass up the slope. I still can't get their formula, but it may be a simple approximate rule of thumb for small slopes.

It's good you found this reference, because it does quote a coefft of rolling resistance for PU as 0.030 to 0.057 an order of magnitude below your 0.4 Caveat on this is, these figures are quoted by people making wheels for material handling carts pushed by hand at up to 3 mph. I have read that, this figure will increase at higher speeds.

In your formula, the first term seems to be rolling resistance on a horizontal surface and the second term the rolling resistance on a slope. So I'd expect only one or the other to appear, not both?

My own view is that I would calculate from first principles and base my calculations on a clearly labelled diagram.

Since we now can reduce our rolling resistance by a factor of 20 (10 for the coefft and 2 for the double accounting), Baluncore is probably right to concentrate on gravity as the dominant force, particularly if you do intend to go up 12 degree slopes ! Rolling resistance and aerodynamic drag are probably more relevant if you go faster on the flat.

Re. Baluncore's, power discrepancy not yet explained, you can't fudge gravity (like you might for rolling resistance or drag) so B has to be right. My guess is that you can do 12 mph OR climb at 12 degrees, not both at the same time. Since mathematically you can climb almost any slope if you go slow enough, that is more like a statement about the motors and fixed gear combination.

Although you need to concentrate on the first two terms, I shall be interested to find out what this fourth term is.
Edit: I should have had another look B4 throwing in that last comment. Obviously it is force to accelerate. I just hadn't looked closely.
 
  • #13
Hi fellas,
Merlin3189 said:
It's good you found this reference
Yes sir I've found that link as you mentioned earlier, yet I found some other link where they mentioned 0.2-0.4 and got lost in it.(I'm searching that resource yet unable to find it...) Its your point that brought me back on track.

Merlin3189 said:
In your formula, the first term seems to be rolling resistance on a horizontal surface and the second term the rolling resistance on a slope. So I'd expect only one or the other to appear, not both?
About the first and second term co-existing, when the inclination is not there, Sin( 0° )=0 and NO GRADIENT FORCE WILL EXIST. Yet when inclination is there, the corresponding force will be there. So I guess that they will co-exist when inclination is there and not when there is no inclination. (Wild guess still)

Merlin3189 said:
I shall be interested to find out what this fourth term is
My bad not explaining things properly about that. Its force due to inertia which is sum of linear and angular inertia.

With modified coefficient of rolling friction around 0.02 , gradient with sin(theta) and inertia formulae's (the mysterious fourth term), I came to an final draft of around 0.529KW as https://drive.google.com/open?id=1wQ_yfqQGqQnqV6235KeRiT8aDp8oHAyb

Yep they were using motors of around that value only (420W, yet it eventually gets heated on overrun).

CWatters said:
MgSin(theta) is the force of gravity acting down the slope.
As a matter of fact I've not gone through the sin(theta) term in gradient term till now. Out of curiosity, I've applied that and I ended with 0.5W. Anyway I'll go through it properly and I'll get back soon guys. Thank you guys for all your support in solving my doubt. Thanks for your reply