Motor torque required

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I am targetting following : Design a Toy car which can carry 30 Kg object at a speed of 1 Km/Hr to a distance of 1 Km in 1 hour. Car will have wheel diameter of 15 cm.

Can some expert verify my following calculation and help....??

A. Calculation for speed requirement

Diameter of Car wheel = 15 cm = 15 x 10-2 x 10-3 Km = 15 x 10-5 Km

Perimeter of Car whee = 3.14 x 15 x 10-5 Km = 47 x 10-5 Km

i.e. when this 15 cm wheel rotates, it will cover a distance equal to 47 x 10-5 Km.

So how many times the wheel should rotate to cover 1 km distance. Therefore, ( 1 Km x 1 Rev ) / 47 x 10-5 Km = 2127.67 Rev in hour is to be made for 15 cm wheel to travel 1Km distance

i.e. 2127.67 Rev per hour = (2127.67 Rev / ( hr  x 60 min /hr) = 35.46 RPM speed is required for 15 cm wheel to cover 1 Km in 1 hour .................Veri-1

B. Calculation for Torque requirement

Force = Mass x Acceleration

Acceleration = Speed / Time

We want speed of 1 km/hr and with that speed car should run 1 hour.

therefore acceleration = 1 km/(hr x hr) = 1 km / hr= 1000 m / (36002) sec = 0.28 m / sec2

Now Force = Mass x Acceleration = 30 Kg (Target weight) x 0.28 m / sec2 = 8.4 Kg m/sec2

i.e. 35.46 RPM speed required for 15 cm wheel dia and 8.4 Kg m/sec2 force is required for moving 30 Kg object at 1 km/hr speed through a distance of 1 km in 1 hour. Then what is the torque required ?

Torque calculation = 8.4 Kg m/sec2 x 15 cm x (1 m / 100 cm) = 1.26 N-m / 9.81 = 0.128 Kg - m = 0.128 Kg-m x (100 cm / m ) = 12.8 Kg-Cm Torque to be delivered.

 

Summary : 12.8 Kg-Cm torque should be produced by a motor at 35.46 RPM speed required for a 15 cm dia wheel to move an object of 30 Kg at a distance of 1 Km with a speed of 1 /km/hr in 1 hours.

This topic has a solution.

Viren Chocha.... (what's up, doc ?)

Last Edited: Mon. Dec 26, 2016 - 02:15 PM
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Now think about it.
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Your car will start at 0 km/hr and end up at 1 km/hr.
You will have to make an infinite acceleration to get to your target speed instantly. Otherwise you will not complete the full 1 km in the hour.
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You could make a steady acceleration from 0 to 2km/hr i.e. with an average speed of 1km/hr. But you would need to decelerate instantly to get to your final 1km/hr target speed.
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Think about a real journey. You accelerate to a cruising speed.
If you are a dragster, you burn a lot of fuel and wear out a lot of rubber.
If you are Al Gore, you concentrate on effeciency and wasting the minimum of resources.
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Your professor gave you a harder exercise than you realised.
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David.

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that's why I am asking for liitle help..lol...

Viren Chocha.... (what's up, doc ?)

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A real life vehicle will have rolling resistance. So you will accelerate gently and the rolling resistance will slow younback to your final 1km / hour.
Just invent some losses. Invent an acceleration scheme. Solve for final speed and final time.
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A bit like a regular artillery question. Or even a rocket trajectory question.
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I bet that your class has already done these problems. It is regular maths rather than microcontrollers.
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David.

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Yes, the problem statement lacks some requirements. 30 kg for a toy car? That's a big toy...

 

Anyway, if you assume a conservative acceleration and deceleration time of 30 seconds each, then the average speed will about 0.99 km/h which is close enough.

 

There are many other things missing from the requirements, which probably don't matter for a theoretical exercise. If it is a practical exercise, they should be taken into account.

Bob.

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Lol....toy car will be for my kid who will physically sit and drive... literally..Ok. ... Conversely, if i i have a stepper motor with 10 kg-cm torque..,how much load it could carry....10 kg at 1 cm radius wheel?....right..

Viren Chocha.... (what's up, doc ?)

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One G is 22 mph per sec. So .1G is 2.2 mph per sec. So if the motor has enough hump to accel at .1G for 5 sec you are going 10 mph. Since f=ma, you know m and a, and f will be the motor torque times the wheel radius, which is known, so we can find torque needed. Motor got that much torque? I leave the si to english units as an exercise for a unit converter applet.

 

Imagecraft compiler user

Last Edited: Sun. Dec 25, 2016 - 07:32 PM
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Without any other information you can assume frictional losses (types, bearings etc) as a drag equivalent to 1% of the vehicle weight - i.e. if the vehicle weighs 30kg the drag would be 30*0.01 = 0.3kgF = 2.9 newtons.

 

This also assumes that the speed is low enough that the aerodynamic losses are negligible.

Real world drag for cars is in the range 0.5% - 1.5% on tarmac, bicycles and trains can be a lot lower.  Driving over difficult surfaces (e.g. off-road, sand...) can be much worse. This is referred to as rolling resistance.

 

This can provide the steady state torque and power requirements.  Additional power for acceleration needs to be added to this.

 

Kevin

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An alternative engineering approach is to look at similar examples. Searching for "ride on motor" has a lot of hits, but it is harder to find spec for torque.

 

Even small motors provide a surprising amount of torque. Obviously it depends on gearing, but 12,000 rpm to 36 rpm is a ratio of over 300. 10 kg-cm seems to be in the right ball park.

 

I think a more limiting factor might be power consumption and battery life.

Bob.

This reply has been marked as the solution. 
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Ok...I understand that the calculation I made is fairly valid for a rough cut trial....Thanks a lot everyone.

Viren Chocha.... (what's up, doc ?)