I ran some test with an in-line watt meter on a dual drive and my weight being apx 185 lbs going up a hill that was apx 10% grade
Using 12s Opto Car Esc’s and Carvon V2 direct drive motors with 90mm wheels
Both tests where at apx same speed
12s - 18a total from battery at half throttle
6s - 36a total from battery at full throttle
That should put 10s at apx 27a total
But as previously mentioned, there are many variables.
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83mm currently but might soon be 97mm
Good info to know! thanks!
I don’t think 25a per motors will be enough unfortunately
@Orin635 here is a chart predicting the performance you will get with 25a battery current limit and 80a motor current limit per motor (83mm tire, 3:1 gear reduction, 2 motors, 270kv, 22.8v battery)… top speed looks to be about 19mph (red line, bottom left chart - thrust pounds minus wind drag force) and peak vehicle thrust is roughly 90lbs (yellow line, bottom left chart - vehicle thrust pounds 2 motors):
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So much information right there its very hard to understand it. Is that a website you got that information from?
@Orin635 here are some of the formulas involved in the prediction:
you can start with the ground speed and wheel diameter to get the wheel rpm,
next use the wheel rpm times the gear reduction (wheel teeth / motor teeth = gear reduction) to get the motor rpm.
next use motor rpm / motor kv to get the back emf v,
then use ((pack v * duty %) - bemf v) / winding resistance = motor current
then 60 / (2 * pi * kv) for motor torque in newton meters per motor amp
then motor current times the torque per motor amp for motor torque
multiply the motor torque times the gear reduction for wheel torque
vehicle thrust comes from relationship between the wheel torque and wheel diameter-- (wheel torque in newton meters * 1000mm) / ((1/2) * wheel diameter in mm) = vehicle thrust in newtons per motor
then you can get the mechanical watts from motor torque in newton meters times motor angular velocity in rad/sec
electical wattage is (pack v * duty %) * motor current = electrical watts
battery current is electrical watts divided by pack voltage
you can calculate the ohmic heating from motor current * motor current * winding resistance = wattage copper loss & ohmic motor heating
wind drag force in newtons is (1÷2) * 1.225 kg/m^3 fluid density of air * estimated frontal area in m^2 * (velocity in meters per second * velocity in meters per second) * estimated drag coefficient = wind drag force in newtons
wind drag power/wattage is wind drag force in newtons * meters per second = wind drag power/wattage
ignored is iron loss, esc loss, battery loss, drivetrain loss, and rolling resistance.
So what sort of information does the diagram give me? I dont really understand it
professor_shartsis:
you can start with the ground speed and wheel diameter to get the wheel rpm,
next use the wheel rpm times the gear reduction (wheel teeth / motor teeth = gear reduction) to get the motor rpm.
next use motor rpm / motor kv to get the back emf v,
then use ((pack v * duty %) - bemf v) / winding resistance = motor current
then 60 / (2 * pi * kv) for motor torque in newton meters per motor amp
then motor current times the torque per motor amp for motor torque
multiply the motor torque times the gear reduction for wheel torque
vehicle thrust comes from relationship between the wheel torque and wheel diameter-- (wheel torque in newton meters * 1000mm) / ((1/2) * wheel diameter in mm) = vehicle thrust in newtons per motor
then you can get the mechanical watts from motor torque in newton meters times motor angular velocity in rad/sec
electical wattage is (pack v * duty %) * motor current = electrical watts
battery current is electrical watts divided by pack voltage
you can calculate the ohmic heating from motor current * motor current * winding resistance = wattage copper loss & ohmic motor heating
wind drag force in newtons is (1÷2) * 1.225 kg/m^3 fluid density of air * estimated frontal area in m^2 * (velocity in meters per second * velocity in meters per second) * estimated drag coefficient = wind drag force in newtons
wind drag power/wattage is wind drag force in newtons * meters per second = wind drag power/wattage
ignored is iron loss, esc loss, battery loss, drivetrain loss, and rolling resistance.
and the load is assumed more I guess with the larger amps used
the yellow line on the bottom left chart shows how many pounds of thrust you have at different speeds with those settings.
Hummie:
and rider weight right?
if you want to calculate acceleration in m/s^2 then you need rider weight, but it isn’t a factor in calculating vehicle thrust or wind drag force on flat ground.
so at 20mp/h i’ll have 0 pounds of thrust?
50amps and 12s works good for me doing steep hills. enough. im 155lbs. especially with a pulley setup should be even less an issue
right, notice how the back emf voltage produced by the spinning magnets on the rotors (which opposes the battery voltage - red line top left chart - back emf voltage) increases with speed and surpasses the maximum pwm effective voltage of your battery/controller right below roughly ~20mph (yellow line, top left chart - effective pwm voltage).
There’s no 12s option from diyeboards and if I was making it myself or buying from a battery builder I would just get a higher discharge cells
Is this diagram for a 10s battery? or 6s?
6s – assumed was 3.8v per cell * 6 cells in series = 22.8v battery
Ok so the diagram for what Is for my current setup?
yes-- with 25a battery limit & 80a motor limit per motor.
Ok, i think there was some confusion. The diagram is a mix between my current setup and what I want it to be/upgrade it to
very sorry
