DC Motor

lower the voltage it will turn slower if its DC

1 hp = 746W

400/746=.536 hp

hp = (torque x speed)/63025 in*lb

torque = force x distance

You can measure the generator output with a multi-meter or ammeter.

The smaller the sprocket you can put on the motor output will maximize torque and the larger the sprocket you can mount to the 26" wheel will do the same.

The "gear ratio" is the ratio of the number of teeth on the driving sprocket divided by the number on the driven sprocket. For example, if you have a 22 tooth sprocket chained to a 44 tooth sprocket, the output speed will be 1/2 the input speed and the output torque will be about 2 times the driven torque.

If I read your post correctly, you have 22 teeth (motor) driving 44 teeth (intermediate shaft), then 22 teeth driving 44 teeth (26 in wheel). That's a gear ratio of 1/4, or an output max speed of 375 rpm (i.e. 1500/4), or 2552 ft/min, or about 29 mph. That's the speed with the motor running as fast as it will with 24 volts, delivering no more torque. Depending on friction and the uphill/downhill grade and weight of the vehicle, your actual speed will probably be less than this.

A dc motor acts as a generator, but, as I understand, not a particularly efficient one. This has been visited before on this site:

http://cr4.globalspec.com/thread/63150/Efficiency-of-PM-DC-Motor-Used-as-Generator

you did far more work on this thread than I ever would have

hey,

The weight of the vehicle will be around 400 kg including drivers and weight on the rear will be around 230kgs.We will be using compound gear system for transmission where the intermediate gears will be of 22 teeth and 44 teeth.We have a gear cassette at the rear wheel and I am not able to select what type of sprocket to use for the to get the torque, so that the motor is able to pull the vehicle.Can u help me out with this?

I don't see you moving 400KG with that motor gearing setup!

Figure out how much of a grade (degrees) you need to climb. Force needed = weight x tangent (grade angle). Output Torque = Force needed x wheel radius. Motor torque = Output Torque x gear ratio.

I am assuming that you have two gears on your "intermediate shaft", a driven gear and a driving gear. If there is only 1 gear (idler gear) it won't do anything for your gear ratio.

A good thing about driving this with a DC motor is that as the motor slows down, it draws more current and it's torque increases. It will slow down until the torque produced equals the torque needed or until it stalls.

I hope this helps.

1. Divide output gear number of teeth by drive gear number of teeth, that's the torque ratio.
2. Divide drive gear number of teeth by output gear number of teeth, that's the speed ratio.
3. When used as a generator @ 1500rpm, it will give you slightly lees than the motor's rated current and voltage.

thats one fast tricycle! use this

It is indeed, but he was asking the basic facts, not the feasibility. Very nice headwear by the way.

I do not know, we are not fund of it anymore around here

Urban Triker Gang.....

WOW!!

They are GANGSTA!!!

Motor cross always way better round here

Let me guess....The guy with the 5 wheeler has a hitch for his horse trailer, right?

Great collection of Trike photos, thanks for sharing!

Forget the brushless motor, use a series wound universal DC motor. The only limit to speed then, would be the ability to keep your feet on the pedals and the commutator bars leaving.

400 watts is the power available at the shaft. The motor can be used as a generator, and in that case yo u have to know the output power, Ra, Rsh values etc.

Telling me the intermediate shaft with 22 teeth sprocket and 44 teeth sprocket. doesn't help in any way. You need to know and tell me what number of teeth the drive (motor) sprocket has and the the number of teeth the driven (back wheel) sprocket has. IF the drive chain is currently connected to the 22 tooth sprocket and the coupled 44 tooth sprocket is attached to the driven sprocket on the back wheel you can increase torque by swapping the chains over so that the chain from the motor goes to the 44 tooth sprocket.

All i have said relates to your statement that the intermediate shaft has 22T and 44T sprockets on the shaft. I am calling this coupled, as in 'joined together'.

Do you know about the old bicycles called Penny Farthings? they had a very large front wheel because one revolution of the pedal gave a long distance of travel on the ground. If the wheel had been smaller the pedaller would have to do many revolutions to cover the same amount of ground. The modern bicycle has smaller wheels that are SPED UP by using sprockets and chain. A small driven sprocket is fed by a large driving sprocket. Once you put an electric motor in as the driver you have revs to spare but not as much torque as a pair of legs and so the driving sprocket has to be small and the driven one has to be large. i.e. SWAPPED AROUND.

Hope this helps

Jim

Your DC motor used as a generator should generate a little bit less than 24 volts at 1500rpm. Most of the voltage applied to a motor without a load is opposed by the back emf as the motor acts as a generator, the remainder being I*R drop.

If you are planning to use the motor for regenerative braking, be aware that the internal resistance of large batteries (lead-acid for instance) is very low, which is why they can supply large currents. If the motor is connected directly to the battery without some means of limiting the current and the motor is turning fast enough to generate back emf in excess of the battery voltage, the current flowing back into the battery could be more than the motor windings can handle.