torque calculation...

Discussion in 'Mechanical | Automobile | Aeronautics' started by vvishwaskumar, Nov 19, 2009.

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hi
i have a bicycle having a load of 100 kg acting on it. I have to calculate the starting torque, to propell the bicycle, either on the wheel or on to the pedal(crank) considering the rolling friction....please give some numerical explaination...please

gohm Moderator

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You are missing information. gear ratio? wheel diameter? length of crank arm? With your info you can only deduce the load is 50kg per wheel if two wheeled. You will take this 50kg weight at the drive wheel and multiply by the ratio of drive wheel size vrs. crank arm. Then multiply by the ratio of the front vrs. rear sprockets.

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hello gohm..
gear ratio = front sprocket/rear sprocket =2
length of crank arm= 0.17 metres
so effort required = 50*2*(0.33\0.17) = 194.14 kg
torque on pedal = 194.14*0.17*9.81 = 323.76 N-m
is it possible to have this much of torque..?
also a 150 cc bike is normally have 14 to 16 N-m torque, can it compare with bicycle calculation..? give me detailed explaination please

BigOhmCertified CEan

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100 kg? Gohm, tell your little brother to quit riding on the handlebars?:roll:

There are a bunch of ways of looking at this, but one you might consider is just in terms of work. Think of the 100 kg bike. It's weight is 980 N and you might guess the rolling friction to be 0.05 with good tires and a good level surface. So, every meter the bike goes, you have 49 Nm of work. So work backwards and see how many radians your pedal crank has to go through to move the bike one meter. Throw in a 0.70 efficiency factor for the chain, bearings, etc etc. Then just set (Torque)(radians)(0.70) = 49 Nm.

Sorry for not using math symbols. I'm a newbie and can't find them.

Rohan_sKCertified CEan

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Gohm, dont you think that the Total Weight of the bike ie 100 kg must be considered while calcualting the Work done or hte Effort needed.

The bike weighs 100 kg, so each wheel takes up 50 kg. But the force required to drive will be the total Inertia force ie 100 kg plus the frictional resistance ( here 0.05).

So, Total Effort ( without the sporcket ratios, etc) = (W+f)9.81 = (100 + 0.05(100))(9.81). You can then multiply the sprocket wheels, crank arm ratios and get the Final Effort and Torque required.

Correct me if I am wrong.

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thanks rohan, but how this much torque is possible..think and
also a 150 cc bike is normally have 14 to 16 N-m torque, can it compare with bicycle calculation..? 100kg=bicycle weight(15kg) + rider weight(85kg)

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hey guys help me for calculation......please

rpowerCertified CEan

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give the coefficient of friction.
You want to calculate starting torque that means the torque to overcome static friction bw wheel and ground.
static friction(forcew) = u(coeff of friction) x N (weight normal reaction)
torque at wheels=forcew x radius of wheel.
divide this by 2(gear ratio)
u'll get the torque applied by rider to accelerate the bike from 0.

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thanx rpower...
i am getting from this .torque on pedal = 8 Nm and effort applied on pedal=5 kg..and it looks feasible because 150cc bike normally have 15 Nm of torque.
also can we compare the bicycle torque with that of bike..in terms of their mechanics....? reply please

rpowerCertified CEan

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I do not understand your question compare torque of bicycle with motorbike "in terms of mechanics" !!

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a 150 cc bike is normally have 14 to 16 N-m torque, can it compare with bicycycle torque...?
also we are using a chain drive for transmitting torque from wheel to pedal...so is there no any consideration for tension on sides and moment of inertia....?

rpowerCertified CEan

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in a bike or car there are several inertial loses. innertial loses in transmission , then in differential (in cars), in running engine components.
e.g Torque(clutch)=torque(engine)- Ia
where I is moment of inertia of engine cmponents and a angular acceleration

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sorry

but in calculating torque do u not think that, as the power is transmitted by chain drive from front to rear sprocket, there is also a consideration of tension in the chain...?

rpowerCertified CEan

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Ofcourse, tensions on tight side (upper) and slack side (down) of the chain drive are to be taken into account.

[T(tight) - T(slack)] velocity = Power

gohm Moderator

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As far as the weight issue, it totally depends on if you are only condsidering torque at the drive wheel. Energy loss would be used in the calculation to equate a procise calculation however this problem does not appear to need to factor that based on the info provided by the OP. Sounds like a class theory problem.

sachin121Certified CEan

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hi have vehical of four wheels having 150mm dimeter. i need to transport load of 50kg with velocity 3m/s then which motor should i select .please send me motor specification

sophia9revCertified CEan

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some great tip guys.. ....my dads a mechanical engineering but heslearning to use the forums

beyonbalaCertified CEan

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I don't understand how did u arrive at 15Nm ..
Also when searched for static friction co efficient between wheel and tyre, it is found to be 0.7-0.8.
what is that 0.05?

techfreshCertified CEan

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