Momentum and inertia - study assistance

I think I have a good seat of pants understanding of momentum and inertia. I do NOT have a good understanding of how to do the calculations. It has been too many decades since my physics classes.

For example
with a motor developing 100 ft lb of torque what is the time to accelerate a vehicle weighing 4500 lbs from 0 to 60 mph on level road.

Where can I find a good study guide? Ideally using only high school algebra and not calculus.

Thanks
Dan Bentler

Are you trying to rotate the car or accelerate the car?

I would divide the 100 ft/lbf by the radius of the wheel to get the force applied. If the radius of the wheel has a 1 ft radius then it applied 100 lbf to the ground. 1 lbf / 1 lbm = 1g. So divide the 100 lbf / 4500 lbm to get 0.0222 g which is not very high. Now you can mulitply 0.022222g by 32.174 ft/sec^2 per g to get 0.715 ft/sec^2

Peter
Thanks - I obviously did not set up the problem statement correctly. I want the vehicle weighing 4500 lb to accelerate from 0 to 60 in a straight line on a level road. Vehicle driven with a motor developing 100 ft lb torque. Solve for time.

A guy in another email gave me a formula I have a hard time making out because email is not easy to use to show formulae. His answer was 50 seconds (with torque of 75 ft lb) with torque at 150% of 75 ft lb his answer was 40 second.

I would divide the 100 ft/lbf by the radius of the wheel to get the force applied. If the radius of the wheel has a 1 ft radius then it applied 100 lbf to the ground. 1 lbf / 1 lbm = 1g.
I understand so far.

So divide the 100 lbf / 4500 lbm to get 0.0222 g which is not very high. Now you can mulitply 0.022222g by 32.174 ft/sec^2 per g to get 0.715 ft/sec^2
32.174 is gravitational constant is it not? Do I really need to put it in terms of that? I know it is common to express in terms of g forces but I find that confusing and translate to lb force (entuitively or at least try to).

Its the pits when you dont work with the fine details of physics all the time and have all the formulae memorized.

Dan Bentler

As you've mentioned inertia - what inertia are you wishing to take into account ?

Dan, the formula for linear motion is F(force)=m(mass)x a(acceleration). Unfortunately, 4500lb is weight not mass so you have to divide it by the acceleration of gravity (32ft/sec^2) to get mass.

You also don't have force but torque instead. So, dividing torque by the radius of the driving wheel will give you linear force.

Solving for acceleration gives you a=F/m.

Someone else is going to have to give you the elapsed time once you have the acceleration rate. That would be the 2nd integral of accel which I've long since forgotten how to do.

leitmotif said:

Peter
A guy in another email gave me a formula I have a hard time making out because email is not easy to use to show formulae. His answer was 50 seconds (with torque of 75 ft lb) with torque at 150% of 75 ft lb his answer was 40 second.

Click to expand...

I have no idea how he computed that.

You need to know the torque curves to and a bit of calculus to compute that because the torque is always changing as a function of rpm/velocity. You can assume the acceleration is a constant 0.715 ft/sec^2 but you know the torque isn't constant. Do you have data for the torque curve? Then we can compute the acceleration for every RPM or velocity.

If you assume linear acceleration then you can divide the velocity by acceleration. Using my numbers 88ft/sec divided by 0.715ft/sec^2=123 seconds.

Dick - I'm an SI units person and weight is commonly referred to in Kg (this is actually the mass). If someone really told you the weight, they would have to give it to you in Newtons.

I'm sure Dan really meant the mass of the vehicle was 4500 lb

To carry Dick's analysis one step further.

Draw a graph of velocity vs time. For now, make it a straight line.

V |

  |        ___________________

  |       *

  |      *

  |     *

  |    *

  |   *

  |  *

  | *

  |*_______________________________t

The slope of the velocity line is the acceleration. That's where calculus comes in. The derivative of the function describing a curve is equal to the slope. Acceleration is the derivative of velocity.
You know that the slope of a straight line is the change in the vertical axis value divided by the corresponding change in the horizontal axis value.
Since the slope is equal to acceleration you can say:

a = Change in Velocity / Elapsed time

Solve for time.

A real world system can't behave quite like that, but it will establish a minimum time to go from zero to 60.

http://www.eaton.com/EatonCom/Searc...for+Adjustable+Frequency+Drives&submit=Submit

The first item, the application guide, will help.

For a design B motor I usually use average accelerating torque is 1.66 times full load torque.

Is there any gearing between the motor and the wheels via a differential for example ?

OK maybe this time I will get the problem statement set correctly.

Truck on a flat level straight road.
Driven with 50 HP motor 3600 RPM with torque of 75 ft lb.
Motor is supplied with a vector VFD supplied by battery.
Motor direct coupled to driveline
Driveline RPM at 60 mph is 3600
Assume rear end ratio of 4:1
Rear axle and tire RPM is 900 at 60 mph
Scale weight of truck is 4500
Motor torque is constant since operating at baseline or less.
Vehicle is constant torque load
Disregard wind resistance
CALCULATE TIME to accelerate from standing stop to 60 mph.

I think I have the formula figured out - it is
WK (squared) = Weight X [FPM / (6.28 X shaft RPM)quant squared]
That should result in 307

THEN
Time = WK/\2 x delta RPM / 308 / Tq
t = 307 x 3600 / 308 / 75 = 47.8 sec

308 is a conversion factor involving torque.

Have more studying to do.
Will look at the Eaton referance - thanx Tom.

Dan Bentler

Here's my SI unit calculations.

leitmotif said:

Driven with 50 HP motor 3600 RPM with torque of 75 ft lb.

Click to expand...

leitmotif said:

Motor torque is constant since operating at baseline or less.

Click to expand...

Yes, but you can get a lot more than the rated torque for a short period of time. See my chart.

leitmotif said:

Vehicle is constant torque load
Disregard wind resistance

Click to expand...

I'm presuming you mentioned this because??

My experience at the salt flats is just the opposite.

Afer reading the new problem statement I realized that you are making this tougher than it need to be. As LD, our friend from the UK, indicates this is actaully a problem in linear acceleration, not angular momentum. His methodology is correct.

Hint: watch your units if you are doing this in standard US units- mass is not the same as weight!

And note Vaughn's comments about torque - you can set a VFD to a wide range of accelleration torques. The 1.66 figure I used was for across the line starting.

Milldrone Throwing out wind resistance was to make the issue simpler to solve. I wanted simpler to make my lernin simpler. In real world yes wind resistance is an issue above 30 or so At Bonneville a much bigger issue since wind resistance is a square factor of relative velocity.

I wanted to know what the time was without motor overload ie slowest or worst case. In reality the peak torque and HP will be used but hopefully for short periods such that average load is 95 or so % or less.

LDAR thanks for the calcs. I intend to copy them to a file for referance. What did you use for a rolling resistance value and where did you get it from? Do you really have a value specific to a 56 Chev pickup? If an assumed or generic commonly used value well who am I to criticize. You came up with the calcs in an hour. I have to dig it out of books because I am definitely in a "use it or lose it situation".

Dan Bentler