How much torque is needed to sustain a 3500 lb car at 70 mph?

[mender]

My desire is to have a 5.3 litre 4x4 4 door chevy colorado that at its best is
a 17 mpg hwy 13 city mpg truck that would have
5 to 8 mpg increase in fuel milage .
At 4.00 a gallon for gas and 4.50 for diesel
a savings of around 50 bucks every 100 miles on the hwy.
That would be really neet to have plus save fuel.

That would be a really neat trick considering it only costs you about $23 in fuel now to go 100 miles! The best you can expect would be more like $7 every 100 miles if everything works as planned.

As far as understanding the concept, I'm quite sure that all participating are fully versed on what you're proposing. Good luck, and be sure to post when you get finished!

 

[mender]

Thanks bob I will be back one day for some help on gear ratios
to get the rpm,s 0f the mini engine correct.

If your truck has the 3.73 rear gears, your driveshaft will be turning at 2780 rpm at 70 mph, meaning that you'll need to gear down your little engine(s) by about 20%.

 

[fishredeemer]

And here is an example.

My car takes an average of 8.32 seconds (four runs, two in each direction in zero wind, 3000 ft ASL and -1 C) to decelerate from 115 kph to 105 kph (Canadian car - easier to use Kph than mph because of the size of the markings). Converting that to mph, a difference of 6.21 mph or 9.11 ft/second.

Dividing 9.11 ft/second by 8.32 seconds gives a deceleration rate of 1.09 ft/sec^2, which is 0.034 g. Multiply that by the weight of the vehicle (3700 lbs in this case) and we find that it takes 126 ft.lbs of force to maintain roughly 70 mph.

To find the hp needed, convert 70 mph to ft/second (70mph/60mph x 88 ft/second = 102.7 ft/second. Multiply the force by the speed (126 ft.lbs x 102.7 ft/second), then divide that by 550 ft.lbs/second and the result is that my car requires 23.5 hp at about 70 mph.

This is some good stuff you guys are putting up thanks for the effort you have given
grately appreciated. As you have fiqured out I have no education to be able to do this calculations . I ve had lots of training in the auto industry and years of experince in how
things work . Drop out 12 grade ged passed . I am the kind a person that loves to make improvements behond the normally exceptible operation of mechanical things.
Thats just kinda a hobby I enjoy with trial and error are failure to success.

These formula's and measurments that was calculated to figure the hp measurement
is very educating and I believe to be very accurate. I m sure the for me to learn how to do that math would be nearly but not totally impossible. Just from passed experience and
and training i estimated that it would be 27 to 30. I could have easyly missed it by far greater numbers .Thats why you guys build rockets and such and I just repair autos
there's not much room for a educated guess .
Thanks a bunch.

I now know where a good mechanical engineer can be found.

 

[Bob S]

A typical internal combustion engine has maximum fuel efficiency at roughly 35% of max RPM and 80% of maximum torque. So to get 35 HP at maximum efficiency will require roughly an engine rated at 125 HP. Look at the BSFC (brake specific fuel consumption map) of the engine to find the optimum torque and RPM.

Bob S

 

[fishredeemer]

If your truck has the 3.73 rear gears, your driveshaft will be turning at 2780 rpm at 70 mph, meaning that you'll need to gear down your little engine(s) by about 20%.

Thanks bob well appreciated info.