[kennybobby]

fissiks

Hi Could somebody help me please?

Could someone tell me what voltage DC motor I would need to propel a 1400kg car to 70mph and the explanation behind it please too.


Thanks


Jon

Sent from my F8331 using Tapatalk

How quickly do you need to accelerate up to the 70 mph, e.g. 0 to 70 in xx seconds? For a rough estimate this determines the power rating of your motor.

So first i'm gonna convert some units; 1400 kg is about 3100 lbs, divide this Weight by gravity, g, to get mass, m; and 70 mph is a velocity, v, of about 103 ft/sec.

Now the kinetic energy of motion at 70 mph is (.5mv^2) = 510682 lb-ft

For a first order power estimate take the kinetic energy divided by your acceleration time, let's assume 0 to 70 in 13 seconds.

This gives a power of 39283 lb-ft/sec, a horse can pull 550 pounds over a distance of one foot in one second, so divide by 550 to get about 72 horsepower required for acceleration.

Your specific question of motor DC voltage can not be answered without additional information, such as the size of your battery energy storage system available to supply power for your vehicle. The voltage and current rating of your battery pack is the deciding and/or limiting factor for the motor selection. But we can take a look at some variables involved.

72 hp multiplied by 746 gives you power in Watts, about 54000 W.

Now this power can be supplied as Volts times Amps = 54000.

If you have 1000 Amps available, then you only need 54 Volts, or
if you have 500 Amps available, then you only need 108 Volts, or
if you have 400 Amps available, then you only need 135 Volts, or ....etc.

See how this is going? there is no unique voltage solution until you select the limits for your acceleration time and your battery pack current. Then you can determine how many cells you need to make the voltage necessary, and then you can specify the motor winding voltage rating for the peak power that you need.

 

[kennybobby]

No, i'm just a ******* with a libary card...

 

[kennybobby]

An Intuitive Approach to Fissiks and Mechanics

Take a look at your Craftsman torque wrench and see the units of torque: ft-lbs.

Every mechanic knows what a 5 lb sledge hammer feels like in the hand, but who has an intuitive feel for 15 Newtons?

So take a ratchet wrench that's 12 inches (1 ft) long and pull the handle with 5 or 10 or 50 lbs--that's gonna be 5,10 and 50 ft-lbs of torque.

How many mechanics have a 1 meter long ratchet in their toolbox? and what's the 15 N pull force feel like to generate 15 m-N of torque?

So here is my intuitive reason why the correct units for torque is ft-lbs or m-N.

Torque is a vector cross product defined as R x F = |R|*|F|*sin(theta).
Reference: https://en.wikipedia.org/wiki/Torque

The equation for Work (Work is equal to the change in kinetic energy), which is a vector dot product or scalar with the units of energy.
Work = F . L = |F|*|L|*cos(theta)
Reference: https://en.wikipedia.org/wiki/Work_(physics)#Mathematical_calculation
https://en.wikipedia.org/wiki/Work_(physics)#Work–energy_principle

Both torque and work appear to have similar units, but confusion can be avoided by following the order of the vector equations: ft-lb for torque and lb-ft for work or energy.

Or in metric units: m-N for torque, and N-m for work or energy (N-m is Joules in SI units).