How to Convert the Mechanical Loads to Electrical Energy

Almost impossible.

If you are travelling on a flat surface the only energy used is to overcome friction and wind resistance, and, these are both things which can vary hugely from one design to another.

Of course you could easily do the calculations to work out how much energy is used in acceleration and then wasted in braking, or, to get from a low level to a higher level. It would then be a simple matter to convert these quantities to electrical energy, but, you would be ignoring the biggest portion of the energy consumed.

For a simple comparison of engines/motors:-

1 horsepower = 745.7 Watts

Assuming DC: Watts(dc) = volts x amps.

You could measure the volts and amps with a multi-meter or:

watts/746 = hp

hp = (torque x speed)/63025

measure your wheel Ø"

time how long it takes to go a known distance

calculate rpm

solve for torque

OR

find the load and calculate torque based on wheel Ø

time the speed

solve for HP

measure voltage and solve for amps.

I can give it to you in SAE units for AC induction motors, you are on your own for conversion, sorry.

Time to Accelerate a Load:

Seconds = WK² (in lf-ft²) x Speed Change (in RPM) / 308 x AAT (Average Accelerating Torque in lb.-ft.)

AAT for a motor = [(FLT + BDT)/2] + BDT + LRT / 3,

Where BDT = Break Down Torque, FLT = Full Load Torque, LRT = Locked Rotor Torque

WK² = inertia of the motor + [Inertial of the load x load RPM²/Motor RPM²]

Have fun!

It is not hard if you look up some basic info and do a few simple sums for each of the factors - rolling resistance, slope, acceleration, drag and efficiency.

Use Excel or the like and perhaps write some macro's to do the math for each part of the problem and you will get a very realistic answer. In metric units then:

Rolling resistance is about 0.02 x mass (kg), but convert to N, multiply by g (=9.8).

Aero drag is frontal area(m2) x Cd (about 0.4) x Velocity^2 divided by 2

^2 means velocity squared or velocity x velocity, units for velocity are m/s2)

Slope force = M x sine angle (careful as computers often use radians and not degrees for angle) - and multiple that by g to convert Mass units to Force units.

Acceleration force = Mass (kg) x acceleration(m/s2). Answer is already in N

Add all those forces up and multiply by velocity to get power at each speed

Of course 1 m/sec x 1 N = 1 Watt, and 746 Watt = 1 HP

And if you are sizing for a battery. Energy = power x time x 1/efficiency.

Efficiency is about 80% so battery output will be about 1.25 times power required at the wheel.

Energy is usually quoted as kWh so if you have been working the problem in watt seconds for consistency divide by 1000 and multiply by 3600 to get kWh.

Not so hard really, just a lot of little bits.

"Aero drag is frontal area(m2) x Cd (about 0.4) x Velocity^2 divided by 2"

Is the fluid (air) density missing here?

You are correct - it is.

All previous formulae is useful, just be sure to take into account all effort you need to overcome, wheels/road, internal inefficiency (2%, internal frictions), maximum road slope, total maximum weight, and if several speeds are available, check all (torque & power) for all the speed ranges, resistance to wind, electrical light you may install, etc. Good luck.

A great deal depends on what you need the information for.

Is it to design the size an electric motor or the size of a battery or just the running time. Is total weight of the finished two-wheeler critical. Is load carrying required. Is speed more important....???

All these questions will have to be answered in order to establish the power-to-weight ratio require in order to focus on which formula to use to produce the sizes etc, to select the appropriate components.

Given the state of power electronics is it possible the transportation cycle efficiency can be significantly different than steady state power efficiency?

An example would be the effect regenerative braking could have on overall efficiency of the transportation cycle; such as a trip to the store.

Also; wouldn't the efficiency at steady state will be somewhat different than that in acceleration because of the relationship of current to resistive losses?