Climbing hills with as little energy as possible.

[jawnn]

I have been talking to people in town and they all approach this problem from the desire for speed. Obviously I am going to have to do the experiments my self. So now I need a grant to fund this wild idea: ‘how to climb a 16% grade with as much weight as 450 lbs with as little energy as possible’.

I think it’s just a matter of gearing a 24-volt motor down to 60 RPMs and tying it into the drive train. But every one wants me to use a bigger motor and bigger batteries to increase the range and power. Well maybe I could find some one to sponsor a contest, not that it would be a popular concept.

If any one has a realistic reason why this can’t be done I would be interested, but only if they understand that I am not looking for more than 2 or 3 mph.

What I realy need to know is what is the minamum rpms a hlf hp motor will do and still have at 'least' a quarter hp?

[spinningmagnets]

To climb with very high torque efficiently, I would start with high motor RPM's. If you are talking about a 26" mountain bike (MTB) I would also consider using a 24" rear wheel.

to get 60 wheel-RPM's on a steep grade, while having high motor-RPM's (3,000-RPM's?) you will need a significant gear-down (50:1 total range). I would also start with 48V unless you were using one of the high-kV RC motors (which can be quite small) They can provide very high RPM's at lower voltages.

There are several builds that have installed a free-wheeling crank with a double-sprocket. This allows the motor to drive the Bottom Bracket (BB) which is the pedal-axle. In this way the motor can also use the bikes rear-axle gears. Cyclone and Elation sell these type of kits, and many enthusiasts have upgraded certain parts to make them stronger and more powerful.

Many others are adding a large sprocket to the rear-wheel disc-brake flange to make a Left-Side-Drive (LSD). RC motors are available in small sizes that spin to very high RPM's, so there has been some recent work building one-stage and two-stage drives as a gear-down (in addition to the final drive chain to the wheel).

Both configurations will be found in the "non-hub" section.

You also may consider a NiCD battery pack. Lihium is very good but more expensive, LiPo will be the lightest, and of course the inexpensive Sealed Lead Acid (SLA) will be bulky and heavy. NiCD can also easily provide high discharge amps. Best of luck, and have fun...

[Drunkskunk]

Give me a big enough leaver, and I can move the world.


You will want that motor its self spinning as fast as it can, drop the spead by gearing for the most power and most efficancy.
the further a motor is off it's peak RPMs, the less efficently it's running and the hotter it will get.

It takes 1 horsepower (750 watts) to lift 1 ton 1 foot in 1 minute. Moving a 450 pound weight up a 16% grade at 3 mph can be converted to an equasion based off that information. It takes some Trig. Can't do that in my head just now, but TD posted a chart with various weights, speeds, and inclines and the energy needed for each a few months back.

[Drunkskunk]

Found it.
http://endless-sphere.com/forums/viewtopic.php?f=2&t=10039&p=154820#p154820


If I understand it right, thats taking into account wind and rolling resistance already

[dogman]

A big lever lets a small force move the object. But laws of physics will rule that it will still take a minimum quantity of energy to get the specified weight to the specified height.

The long lever still helps, allowing the smaller motor to do it, possibly with less wasted energy in friction and heat.

Of course pedaling more, like you do when you slow down really changes the ammount of energy taken from the battery. So doing that will get you up the hill with less used from the battery. I do this all the time when I need to extend range. The slower I ride, the larger the porportion of the total that is coming from pedaling is. Then you lose less from wind resistance too, and range gets longer still.

On my 5304 hill climber, the slowest I can ride it is about 5 mph, there is no slower throttle setting. A gearmotor though, might be able to go slower, if you have a trottle with fine enough low settings. Mine is sort of, off- --- and then 5 mph. So something other than a direct drive hubmotor would be needed to do slower gearing. On my pedaling MTB, I can ride as slow as 3 mph, but have no gears lower than that. 22 front 34 rear I think on that gear. Pretty hard to balance under 3 mph too.

[monster]

try the stanna stair lifts forum

[Miles]

I'm not sure you need a grant to learn basic mechanics and motor theory... The only thing that really matters is maximising the total efficiency of your system to climb the hill at the speed you choose. The energy used won't be greatly affected by the speed chosen (within reason).

[swbluto]

I'm not sure you need a grant to learn basic mechanics and motor theory... The only thing that really matters is maximising the total efficiency of your system to climb the hill at the speed you choose. The energy used won't be greatly affected by the speed chosen (within reason).

True. The baseline is going to be the weight you're lifting and you can't get under that (energy required = mass*height*gravitational_acceleration. This is also formally called potential energy. Given a hill of some height, your best bet is to lose some weight.), and any energy required after that is the energy required to push wind out of the way, overcome drive-train losses, feed the heat-producing motor, etc. Since the energy required to push wind out of the way is a function of speed, slower speeds and higher total system efficiency will allow one to get close to the minimum amount of energy required.

For maximizing drive-train efficiency, I think you're probably looking at as high an RPM as practical in a direct drive configuration and geared for "hill climbing" - most direct drive hubs suck for this purpose probably because of their low RPMs relative to their power outputs and the limited amount of practical gearing ratios, so a geared hub and and/or an external motor like an extremely powerful lightweight RC motor are common options. I usually find the hill-climbing "gears" by finding the gear ratio that corresponds to maximizing speed in my simulator (It optimizes it for me), and play with gear ratios higher than that that reduce lower heat and reduce the speed to something acceptable. Also, higher voltages (within reason) will be relatively more efficient, assuming the motor is geared down correspondingly.

(The difference in the energy needed to go 3 mph versus 10 mph isn't really significant. Actually, it might actually take more energy to go 2-3 mph because you may need greater reduction stages which entails increased drive-train losses and that may not offset the increase in air resistance with 10 mph.)

[dogman]

To go slower than 5 mph needs gears. But I ride at 5 mph up hills on DD motor all the time, on very little energy since I'm using the lowest possible throttle setting. To haul a lot of weight though would take more energy, that minimum discussed above. You may, however run into motor heating issues if the motor chosen doesn't have air cooling, so chain drive, or stokemonkey type setups might be best. It just depends on how long the hill is, whether motor heat becomes an issue. My fusin, similar to a bafang hubmotor gets very hot in just a couple miles when climbing at 13%.

[chuck]

At a velocity of 3 mph you will need about .6 hp.

Weight = 450 lbs
slope = 16/100, using a^2 + b^2 = c^2 , skip the trig, use 16/101.27
velocity = 3 mph or 4.4 ft/sec
frontal area about 7.5 square feet

The force of drag is 1/2 (air density)(coefficient of drag)(Frontal area)(velocity^2)
Fd = 1/2(.00273)(.9)( 7.5)(4.4^2) = .18 lbs

The force of rolling on my bike is about 3 lbs using maxxis hookworm tires at 40 psi with about 450 lbs of weight
Fr = 3 lbs

The force of the hill is weight * corrected slope or 450 * 16/101.27
Fh = 71 lbs

Add the 3 forces to find the total force
Total force = 74.18 lbs

To find the power required at the pavement multiply by the velocity
74.18 lbs * 4.4 ft/sec = 326.4 ftlbs/s

1 horsepower = 550 ftlbs/sec

326.4/550 = .6 horsepower

To find the gear ratio.

I have a scott 24 volt motor, 1 hp at 3000 rpm peak effiency.

Use a 26 inch mountain bike wheel. Its diameter is 26 inches/12 inches,,,to get feet,,, = 2.167 feet
Its circumference is 2.167 * pi = 6.8 feet

at 4.4 feet/sec this wheel must spin at 4.4/6.8 revolutions per second or .646 rps
convert this to rpm by multiplying by 60
Wheel rpm = .646 * 60 = to about 38.8 rpm

The reduction ratio is the motor rpm / the wheel rpm or 3000/38.8
Reduction ratio = 77.36 to one

For a 2 stage reduction take the square root of the reduction ratio or about 8.8 reduction ratio per stage
3/8ths pitch or number 35 chain will handle .6 hp with a 10 tooth sprocket at fairly low rpms, not ideal, but about the only thing available. Sprockets of 96 tooth are commonly available, use the chain for the wheel reduction.

96/10 = 9.6 ratio for the wheel reduction, lets find the motor stage reduction

77.36/9.6 = about 8 to 1

A 3v v-belt will handle this power at above 95% efficiency. Do not use a 3L belt, they both fit on the same size sheave, the L stands for light duty. The v stands for industrial. Lets try a 12 inch sheave to get that 8 to 1 ratio. 12/8 equals a motor pulley of 1.5 inches, perfectly fine size for a 3v "premium" called a 3vx belt.

Now, we are real close to a final ratio of 77.36 to one

V-belts have a pitch ratio, subtract .05" from each sheave to get the actual reduction. 11.95/1.45 = 8.24 ratio on the motor.
Divide the 8.24 ratio into the total ratio of 77.36 to get the final ratio of the rear drive. 77.36/8.24 = 9.386
ok, I am pretty sure one can find a #35 pitch chain sprocket with 96 teeth at a reasonable price. To find the jack shaft sprocket, divide the wheel sprocket by the remaining ratio, ie 96/9.386 = about a 10 tooth sprocket.

The final ratio is (96/10) * (11.95/1.45) = about 79 to 1..

(79 - 77.36)/77.36 = about 2 % off, reduce your speed of 4.4 by 2% for peak efficiency,

Don't ask me to calculate the efficiencies of these drives, but, the chain will probably be about 90% efficient, the belt about 95%.

Take the horsepower of .6 and divide by the power transmission efficiency of about 85% or .85.

.6/.85 = to about .7 hp the motor needs to put out. The motor and controller as a combination are about 85% efficient.
.7/.85 = about .83 hp. This is what the battery needs to supply. Convert hp to watts. 1 hp = 746 watts

The battery needs to supply .83 hp * 746 watts = 620 watts

For a 24 volt system the amps are 620 watts/24 volts = to about 26 amps

I use agm batteries that have a reserve capacity of 135 minutes at 25 amps. I use a 24 volt system with 68 ah batteries.

My reserve capacity in hours is 135/60 = 2.25 hours.

At 3 mph * 2.25 hours I will travel 6.75 miles up your hill. The scott motor will be warm, not smoking hot, at this output.

chuck

[swbluto]

At a velocity of 3 mph you will need about .6 hp.

Weight = 450 lbs
slope = 16/100, using a^2 + b^2 = c^2 , skip the trig, use 16/101.27
velocity = 3 mph or 4.4 ft/sec
frontal area about 7.5 square feet

The force of drag is 1/2 (air density)(coefficient of drag)(Frontal area)(velocity^2)
Fd = 1/2(.00273)(.9)( 7.5)(4.4^2) = .18 lbs

The force of rolling on my bike is about 3 lbs using maxxis hookworm tires at 40 psi with about 450 lbs of weight
Fr = 3 lbs

The force of the hill is weight * corrected slope or 450 * 16/101.27
Fh = 71 lbs

Add the 3 forces to find the total force
Total force = 74.18 lbs

To find the power required at the pavement multiply by the velocity
74.18 lbs * 4.4 ft/sec = 326.4 ftlbs/s

1 horsepower = 550 ftlbs/sec

326.4/550 = .6 horsepower

To find the gear ratio.

I have a scott 24 volt motor, 1 hp at 3000 rpm peak effiency.

Use a 26 inch mountain bike wheel. Its diameter is 26 inches/12 inches,,,to get feet,,, = 2.167 feet
Its circumference is 2.167 * pi = 6.8 feet

at 4.4 feet/sec this wheel must spin at 4.4/6.8 revolutions per second or .646 rps
convert this to rpm by multiplying by 60
Wheel rpm = .646 * 60 = to about 38.8 rpm

The reduction ratio is the motor rpm / the wheel rpm or 3000/38.8
Reduction ratio = 77.36 to one

For a 2 stage reduction take the square root of the reduction ratio or about 8.8 reduction ratio per stage
3/8ths pitch or number 35 chain will handle .6 hp with a 10 tooth sprocket at fairly low rpms, not ideal, but about the only thing available. Sprockets of 96 tooth are commonly available, use the chain for the wheel reduction.

96/10 = 9.6 ratio for the wheel reduction, lets find the motor stage reduction

77.36/9.6 = about 8 to 1

A 3v v-belt will handle this power at above 95% efficiency. Do not use a 3L belt, they both fit on the same size sheave, the L stands for light duty. The v stands for industrial. Lets try a 12 inch sheave to get that 8 to 1 ratio. 12/8 equals a motor pulley of 1.5 inches, perfectly fine size for a 3v "premium" called a 3vx belt.

Now, we are real close to a final ratio of 77.36 to one

V-belts have a pitch ratio, subtract .05" from each sheave to get the actual reduction. 11.95/1.45 = 8.24 ratio on the motor.
Divide the 8.24 ratio into the total ratio of 77.36 to get the final ratio of the rear drive. 77.36/8.24 = 9.386
ok, I am pretty sure one can find a #35 pitch chain sprocket with 96 teeth at a reasonable price. To find the jack shaft sprocket, divide the wheel sprocket by the remaining ratio, ie 96/9.386 = about a 10 tooth sprocket.

The final ratio is (96/10) * (11.95/1.45) = about 79 to 1..

(79 - 77.36)/77.36 = about 2 % off, reduce your speed of 4.4 by 2% for peak efficiency,

Don't ask me to calculate the efficiencies of these drives, but, the chain will probably be about 90% efficient, the belt about 95%.

Take the horsepower of .6 and divide by the power transmission efficiency of about 85% or .85.

.6/.85 = to about .7 hp the motor needs to put out. The motor and controller as a combination are about 85% efficient.
.7/.85 = about .83 hp. This is what the battery needs to supply. Convert hp to watts. 1 hp = 746 watts

The battery needs to supply .83 hp * 746 watts = 620 watts

For a 24 volt system the amps are 620 watts/24 volts = to about 26 amps

I use agm batteries that have a reserve capacity of 135 minutes at 25 amps. I use a 24 volt system with 68 ah batteries.

My reserve capacity in hours is 135/60 = 2.25 hours.

At 3 mph * 2.25 hours I will travel 6.75 miles up your hill. The scott motor will be warm, not smoking hot, at this output.

chuck

Lot of free time?

[jawnn]

The best advice so far is to start with a high RPM and keep it there.

Stop thinking speed, this is a trike with trailer and plenty cargo.

60 rpm’s will be the crank speed (for my legs) Possibly 80 with muscle input.

My lowest gear is a 1 to 1.88 ratio (18 sprockets front/ by 34 rear). That makes my 20” tire turn at about 32 rpm’s or 1.89mph? (I did not use a formula for this, just brain storming, so I don’t know that it is correct)

The next gear is 28 sprockets front/ 34 rear = 3 mph

Well maybe a bit slower even is the tires are 85psi, but if I use my legs I may need a speedometer.
.
Will a 500w 24v motor push 400lbs plus the weight of the power system, batteries and all, up the 16% grade at 1.75 mph???

This is not a long hill but there are other hills to deal with.

[dogman]

You may do it that slow on 200 watts. You're in an area where you'll have to do your own experimenting to see what works though. A stokemonkey setup on a trike or a cyclone might do it with low enough gearing.

If you think about it, dump trucks haul huge loads up hills with not that big of engines. But they have at least 8 gears, and the low ones are incredibly low.

The total energy used is set in stone, but the size of the engine providing it isn't.

[swbluto]

The best advice so far is to start with a high RPM and keep it there.

Stop thinking speed, this is a trike with trailer and plenty cargo.

The best advice is to keep MOTOR RPM high and gear it down so that you get LOW WHEEL RPM = LOW SPEED. Hard to understand?

This has all the benefits as explained above, the least of which is high efficiency which is what you'll need for minimizing energy usage.

[jawnn]

Well I am compleetly lost...


Will a 48-volt motor have a lower peak efficiency rpm than a 24-volt at the same 500 watt (or possibly less)???

Maybe I should just gear down to 80 rpm’s and enter the drive train between the crank and the rear sprockets. In fact I could use an even higher rpm (how musch higher is beyond my brain power) if I gear it down at the entry point, but remember that I want to power this with my legs only as much as possible. That’s why I want as small a power system as possible.

Seems to me that the gearing down of the motor will be the hardest part.