How much power do you really need?
For more thrust per RPM you need a larger motor or a lower gear.
This math will not really tell you the size of battery you need nor the voltage needed. All it can do is recommend that your motor be bigger than you think [the larger wattage output maximum of a motor, the easier it will be to reach the thrust potential needed even at the same wattage because amps determine the actual thrust].
For hauling over 400lbs up a 10% grade or steeper, you will need a battery pack that has a big enough voltage to let you accelerate hard enough that the Voltage drops at least 5-8volts. With the lowest voltage above the cutoff point of your controller. If you start with a 48volt pack there will be much less than a 60volt pack even if you have more amp-hours in the cells of the 48v pack.
engineer
a power system for heavy cycle-trucks
Heavy
utility cycles will become very popular as people wake up to the
reality of climate change. It is not easy to understand how to
provide enough low-speed thrust that is needed to climb steep hills
at legal speeds. Bigger motors are not the best solution. Proper gear
to power matching is much more efficient and economical in terms of
mileage.
- Measure the average steepness of the worst hill that you cannot avoid climbing. Use a level and a ruler (metric or decimal), then divide rise into the span to get the grade percentage.
- Total combined weight, plan for at least 400-500lbs (181-226 kg). If you have a cycle-truck or a velomobile that can hold two people, think about at least 600-800lbs (272-362 kg).
a.)
Find a power calculator or find a graph for your motor, like the
bicycle speed calculator at kreuzotter.de that has a setting for
Velomobiles. http://www.kreuzotter.de/english/espeed.htm
b.)
Then a gear ratio calculator, like
the
motor
and gear ratio calculator at Electric Scooter Parts Support Center.
c.)
Take
the RPM speed that your motor will produce at the wattage you will
need (see
a performance graph of the motor).
Calculate a gear reduction ratio you will need
to reduce the speed of the motor to get the maximum thrust. Subtract
a little speed (1-2 mph or 1.6-3.2 kph) for the lack of accuracy.
Find
the peak efficiency R.P.M.s and wattage on the label. When a motor is
pushed past the peak efficiency limits too long, it can overheat.
Brushless motors can produce about twice the power on their label for
a few minutes without overheating too much; to get past the steepest
part of the hill.
the myth of ebike wattage:
How much power do you need for a hill? You can work out approximately how much power you need for a hill :
Watts = (total) weight (kg) x 9.81 x speed (M/sec) x gradient (%/100)
For instance 100kg at 20kmh at 4% (3% hill + 1% w+r) = 100 x 9.81 x 20000 x 0.04/3600 = 218 Watts and
100kg at 28kmh at 5.5% (3.5% hill + 2% w+r) = 420 Watts
That is approximately what you can expect from 250w (200w) and 500w (400w) motors.
Many people confuse torque with power. They are different terms. Motors can have high torque and low speed, or low torque and high speed and yet have similar power output. For instance the two different types of e-bike hub motors, the ungeared and internally geared motors have quite similar power, although the former is a high torque low speed motor and the latter is a low torque high speed motor. Its internal reduction gears reduce the speed of the axle and increase torque in same proportion so that the output of the two is quite similar.
Torque is the force that “wants” to turn the motor. Power is the rate at which work is done. The conversion between power and torque is :
Power (watts) = Torque (NewtonMetres) x Rotation
speed (radians/second)
(Newtons are force : 1 kilogram weight =
9.81 Newtons)
In reverse you can work out the motor torque
required to go up a hill at a given speed and wheel diameter and power output
Torque (NewtonMetres)= Watts/radians per second = Watts x 0.5 x diameter)/(kmh/3.6)
218 watts at 20 kmh on 20″ wheel requires
218×3.6×0.254/20=10 NM torque
420 watts at 28 kmh on 26″ wheel requires
420×3.6×0.330/28=17.8 NM torque
You may have seen a Tour de France cyclist going up a 14% slope at about 30kmh. That cyclist is putting out about 1300 watts! You won’t do as well with any bicycle motor. (But if you do have 500 watt motor and do overtake a cyclist training on a hill be considerate – he or she is working hard, but you are not).
Just as with a car’s motor in any gear there is a speed range where the motor performs most efficiently, while at much lower or higher speeds the motor will either struggle or max out. Electric motors have various performances, just as fueled motors do. An electric bike motor can struggle, or stall, just as car motor in the wrong gear will.
You may find a “performance curve” for an electric motor. Usually there will be several variables (power in, power out, rpm or torque, efficiency, current, voltage) plotted against motor rpm or torque. Efficiency will generally be a maximum in the middle range of motor speeds, and be very low at low speeds. The power out will flatten off and not increase much with increasing motor speed despite the input power increasing. This is because the motor loses efficiency at faster speeds and increasing input power has less effect.
For going up steep hills a motor that has high torque at low speed is better. If the speed going uphill is within the high efficiency range the motor will stable and the bike will go up the hill steadily. The speed going uphill will usually be less than the motor’s most efficient speed. In this case the motor will slow, lose efficiency, and struggle…but if you make enough assistance to the motor you can go up any hill at a steady speed.
52 sprockets / 12 sprockets on the rear= 4.333 ratio? X 60rpm crank speed x (20” x pi=62.8inch circumference) = 16327.987 inch min. / 12 inch = 1360.6655ft/min. x 5280ft x 60min/hr= 15.45 mi/hr
Gear reduction: Calculate wheel speed needed then find ratio of gears:
Wheel diameter x pi x motors peak efficiency RPM ÷ 12”per foot x 60min. ÷ 5280ft per mile = wheel speed without gear reduction ÷ 15mph = ? to 1( gear reduction needed.)
Low voltage = low speed = high torque. However a huge gear reduction could have advantages, like a smaller motor:
 |
| Yes there is a method to calculate HP. HP = V x I x Eff/746 HP = Horse power. We normally use Watts V = Voltage I = Current eff = efficiency of the motor Efficiency = Output Power/Input power I obviously cannot advise you what the efficiency of your motor is, but as an example; Eff = 60/75 eff = 0.8 per unit, so 0.8 x 100 = 80% Feed this data into the above; HP = 240 x 5 x 80/746 HP = 129 |
  |
| 1.6HP = 880 lb-ft/sec. This means you can raise 450# at a rate of 1.96 vertical ft/s. On a 16% grade, that's a vehicle speed of 12.22 ft/s or 8.3 mph. Your 20" diameter wheel has a circumference of 62.8" or 5.24 ft, so your wheel is turning 2.33 revolutions/sec or 140 rpm. 3000 motor rpm divided by 140 is a gear ratio of 21.4:1. |
How to make sprockets
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