Increase Fuel Mileage: Calculate Torque for 3500 lb Car

[fishredeemer]

Please help me and calculate the amount of torque
required to sustain a car ,3500 lb car medium wind drag.
aproxamently 70 mph. I know without wind resistance
you can not figure exact but a close estimate would be greatly appreciated.
Im wanting to add a secondary mini diesel engine 662 cc size. horse power 16 at 3400
torque 25 at 3000 . In theory the extremely small diesel engine assisting large v8 gas
engine will raise fuel miledge considerable.
Thanks

 

[Drakkith]

Why would this increase fuel mileage by alot? What theory are you going off of?

 

[AtomicJoe]

I think he thinks the small engine will be more efficient, different fuel too.

 

[JaredJames]

Why would this increase full mileage at all?

You'd need them to be perfectly synced otherwise they will 'drag' against each other.

 

[fishredeemer]

Why would this increase full mileage at all?

You'd need them to be perfectly synced otherwise they will 'drag' against each other.

It will not need to be perfectly in synced Its a power assist set up I am working on .
a small engine that can support the amount of power it takes at said speed will take way less fuel than a bigger engine . For instance ,a 5.3 lt is 5300 cc a .662 is only 662 cc
If its only taking say 40 ft lb,s of torque to maintain 70 mph . The 662cc can supply most of that .Then the 5.3 will be close to no load whitch will drastically reduce fule useage of the 5.3. The mini diesel engine will have very little waste of fuel .the diesel will be performing at its max and performing the needed torque.
So back to the question is any of you able to answer the question.?

 

[Drakkith]

Is this torque value required before or after the calculations from the gears and wheels and such?

Also, the engine may be able to produce that much torque at that RPM, but that is probably with the gas pedal floored, pouring fuel into it. You'd need to find how much fuel the engines would use to keep your vehicle at that speed, not just the torque required.

 

[fishredeemer]

Is this torque value required before or after the calculations from the gears and wheels and such?

Also, the engine may be able to produce that much torque at that RPM, but that is probably with the gas pedal floored, pouring fuel into it. You'd need to find how much fuel the engines would use to keep your vehicle at that speed, not just the torque required.

Running at peak torque is idea for fuel miledge with proper gearing that's what
imtrying to set up for. The amount of fuel used is the purpose of this added engine
It will support the load with better fuel economy.
The amount of fuel for given load will very in many different ways
The peak torque curve is generally the point when engine is most efficeint.
You will not be able to figure fuel mieladge without testing.
This question is a little difficult to answer so thanks Ill need to
go to a automotive performance forum to get a figure .
Thanks

 

[fishredeemer]

Is this torque value required before or after the calculations from the gears and wheels and such?

Also, the engine may be able to produce that much torque at that RPM, but that is probably with the gas pedal floored, pouring fuel into it. You'd need to find how much fuel the engines would use to keep your vehicle at that speed, not just the torque required.

The main reason for question is to find out what percentage of the torque required
will be supported from mini diesel engine.
The engine is to small to apply enough force to accelarate to speed,

 

[Lsos]

16 horsepower (no matter what torque) won't move a car like that at 70mph. We used to have this freaking thing...

...and it had ~24 horsepower and I don't remember it being able to go much above 60mph...although Wikipedia claims it could do 75mph. Nevertheless, it weighed a third as much as your car and is much smaller and I just don't see 16hp getting you there.

I suppose the diesel can assist the gas engine, but keep in mind that a gas engine is most efficient at wide open throttle. You'll have to rein that thing in pretty good, and just the fact that it will have to be on will pretty much negate any fuel savings you can hope to have. That's my theory at least...

 

[mrspeedybob]

Horsepower requirement increases with the cube of speed so go find out what your vehicles top speed is and what rpm the engine is turning at that speed. Then find a dynomometer sheet for your engine and see what horsepower the engine produces at that rpm. From there the math is pretty simple to find the horsepower requred to maintain any given speed. Horsepowere = (torque x rpm)/5252 so the math to get from hp and rpm to torque is pretty simple.

P.S.
Find a track for your top speed test. Don't be stupid and try it on a public road where you will kill yourself and someone else.

 

[fishredeemer]

16 horsepower (no matter what torque) won't move a car like that at 70mph. We used to have this freaking thing...

...and it had ~24 horsepower and I don't remember it being able to go much above 60mph...although Wikipedia claims it could do 75mph. Nevertheless, it weighed a third as much as your car and is much smaller and I just don't see 16hp getting you there.

I suppose the diesel can assist the gas engine, but keep in mind that a gas engine is most efficient at wide open throttle. You'll have to rein that thing in pretty good, and just the fact that it will have to be on will pretty much negate any fuel savings you can hope to have. That's my theory at least...

Power assist ( I am adding small engine to reduce load of big engine) .Small engine at max torque supporting most of the weight and drag Maybe this will help. I thought one of you could get some data .A rough estmate
The varibles to question are endless ,minumum amount of torque to power a car at road speed, in an idea sittuation no flat tires are wheel bearings going out,etc,

 

[fishredeemer]

Horsepower requirement increases with the cube of speed so go find out what your vehicles top speed is and what rpm the engine is turning at that speed. Then find a dynomometer sheet for your engine and see what horsepower the engine produces at that rpm. From there the math is pretty simple to find the horsepower requred to maintain any given speed. Horsepowere = (torque x rpm)/5252 so the math to get from hp and rpm to torque is pretty simple.

P.S.
Find a track for your top speed test. Don't be stupid and try it on a public road where you will kill yourself and someone else.

Fuel Economy, Engine Efficiency & Power
by Rad Davis, Virtual Vairs
Philip Stevens wrote: I'm trying to understand fuel economy and engine efficiency. Does an engine run most efficiently at maximum torque, or maximum horsepower, or where? Also, is the point at which an engine runs most efficiently also the point of maximum fuel economy? I've also been told that manifold vacuum is related in some way to fuel economy. Anyone care to elaborate?

Rad Davis answers: Ah, a technical question! Just what I like.

An engine (no accessories), running at full throttle (we're ignoring power valves and other enrichment gadgets) has its lowest fuel consumption/unit power produced at the torque peak. The horsepower peak is a mathematical artifact of increasing RPM coupled with decreasing efficiency, and is not usually a good place to run for max. economy. Unfortunately, unless you're running a generator, or pump, or something like that, it's not a particularly relevant value, because automobile drivers *do* have engine accessories and power valves (late corvairs, anyway) and *don't* spend much time at full throttle at the torque peak.

OK, now a little quick theory: The reason that economy (efficiency) maximizes at the torque peak is because this is the place where the engine is inhaling the greatest amount of fuel and air into the cylinder which, when burned, makes the peak amount of cylinder pressure and thus the maximum torque.

So the instant you close the throttle some, you're losing efficiency and your fuel economy (expressed per unit power output) goes down. Again, if you're generating electricity or pumping water, this is a bad thing.

But cars are in a variable load environment. You may be going downhill at 60 mph, and only need 1/2 hp to maintain speed. So the engine is horribly inefficient in producing this 1/2 hp, but you're demanding so little compared to what the engine can make that your economy (expressed per unit distance traveled (like MPG)) is extremely high.

If you then complicate things by including parasitic power losses to the cooling fan, vacuum advance, power enrichment, and aerodynamic drag, things get pretty murky, at least from a theoretical standpoint. The good news is that we can make some emperical generalizations:

Assuming that the engine is operating in its powerband (roughly 1200 RPM to 4000 RPM on two-carb corvairs), and throttle is between zero and 80% of travel, the slower the engine is turning, the better fuel economy will be. So upshift early--fuel economy is always best in top gear. This is especially true with a Corvair, because the power consumption of the cooling fan is a cubic function of RPM.
The slower you go (assuming the 1200 rpm/80%/top gear rule) the better fuel economy will be, because aerodynamic drag is the largest single power sink above about 35 mph. I can't find my 110 HP Stock Engine Test Report (lots of stuff still in boxes), but best fuel economy for the late sedan with that engine was something like 32 MPH, according to the dyno numbers.
The less you start and stop, the better fuel economy will be. Every time you start the car, you waste all the kinetic energy as heat by using the brakes. Starting from the stoplight sucks up power (and fuel) replacing that lost energy of motion. Accordingly, a lighter car will get better fuel economy irrespective of any aerodynamic changes. It takes 3200 lb-ft to raise my Greenbrier one foot. It only takes 2300 lb-ft to raise a late coupe that far. Every time you climb a hill this comes into play.
About vacuum gauges: These make more sense on cars with huge engines. If you've got something like a caprice with a 454, the fuel economy of the car is related almost exclusively to the amount of fuel the engine is demanding at that instant, regardless of speed. Our standard joke about our '73 400 CI Pontiac was that it got 13 mpg uphill, downhill, a/c on or off, towing a trailer or not. The reason for this was that the engine could make something like 350 HP. And in normal operation, you used maybe 60 HP. Highway cruise was something like 20 HP. So in typical driving, you could get around with 20% of the engine's total power output. The engine is in a hideously inefficient regime all the time, so all you can do is try to minimize the gross fuel consumption.
Engine vacuum is a good approximation of just how much fuel and air is going into the engine. Vacuum is near zero at full throttle. Higher vacuum=less fuel. So if you use the vacuum gauge as a guide, less is better.

Where this falls down is when it runs into situations when the engine *can* be steady-state loaded near full throttle and the torque peak. In our Pontiac, this would be pulling a trailer climbing Mount Everest at 60 mph in top gear. Fortunately, Corvairs have smaller engines relative to their weight, and weigh less than that Tin Indian did, too.

So if you take something like my Greenbrier (Corvair minivan, if you don't know), and drive down the road with a low rearend ratio (3.27:1) and tall van tires at 65 MPH, you're either at about 60% throttle or you're slowing down. Climbing even a 6% hill on an interstate you get up to 75% throttle or so. So you're in a situation where the engine (as a black box) is starting to be very efficient. Sure enough, the Greenbrier never gets less than 20 MPG on interstate trips, and set a record of 28 MPG driving through the mountains of Tennessee in heavy traffic. By the vacuum gauge, the fuel economy should suck, but in practice, it's better than most turbo corvairs can manage, even though the van is heavier and less aerodynamic.

This is the secret of cars like the Geo (now Chevy) Metro--it doesn't weigh anything, has no frontal area, and has a tiny engine (1 litre). So the little engine still

 

[fishredeemer]

Just put the idea of mine to side .I am 100 percent
sure this will increase mpgp . In an idea road test
what's requred to maintain 70 mph with a 2011 chevy camaro?
Not how much to get to speed just to sustain at speed.
During accelaration The big engine will be doing the work.

So can any of you educated people do the equation?
Its evedent I have not been responded by the one to do this math.
But I will get my answer from the right one. Tuff question huh?
They doint make a rolling test dyno >LOL
They do have on board computers that can give close data.
If I had measurements of resistance it would be easy right??LOL Dont have those
so keep giving imput if you like.
Just to make more sense for you guys ,They have a lot of car makers
with varible cylinder use.To increase mpg.
Switching from 8 cylinders firing too 4 cylinders firing at cruise speed.
4 cylinders verse 8 cylinders = increased fuel mpg
8 cylinders = 5300 cc verse 4 cylinder 2650 cc small eng = better mpg

 

[fishredeemer]

Power assist ( I am adding small engine to reduce load of big engine) .Small engine at max torque supporting most of the weight and drag Maybe this will help. I thought one of you could get some data .A rough estmate
The varibles to question are endless ,minumum amount of torque to power a car at road speed, in an idea sittuation no flat tires are wheel bearings going out,etc,

The small car in picture has very poor areodynamics.
Build a many corvette of same weight and same engine
Then you will have undestanding of question.

 

[mender]

For a typical car that weighs 3500 lbs, rolling resistance accounts for about 11-12 hp and aero drag the remaining 13-14 hp for a total of about 25 hp to maintain 70 mph on a dead flat road once you get up to speed. If you want to check a different speed, the rolling resistance is close to linear (30% increase = 30% more hp required) while aero drag changes as the cube of the change (30% increase = 1.3 x 1.3 x 1.3 = 2.2 times the hp, 30% decrease = 0.7 x 0.7 x 0.7 = 0.34 times the hp)

A better way to do this is to have an engine that is bigger and has a wider efficient rpm range. Ideally, you could build a drive train that can make the 25 hp needed, when running at 2000 rpm and 80-85% load, while being able to provide the maximum reserve power desired at higher rpm. Turbocharging can help fill in the hp demand as well and works very well with a diesel.

Very much like what VW is doing with their TDi Jettas and such; might be cheaper just to buy one and enjoy the fuel mileage without the work.

Don't forget that running your big engine will consume about 1/2 gallon of gas per hour at an idle, which is about the amount needed to produce 4-5 hp with the small engine, plus you're carrying around the extra weight of the normal drive train, etc, so it'll never be as efficient as a dedicated single engine system.

Most modern cars run their engines quite slow to get them into the high load efficient BSFC range and already get pretty good fuel economy for their size. My wife's full-size Buick runs at 1600 rpm on the highway and gets over 25 mpg and has more than enough reserve power. The fuel economy could be improved by up to 40% by reducing the engine size to about 1/3 the present size but would make the car all but useless on a typical drive.

Your quoted article is pretty close but the author isn't quite right on some of the reasons for the effects. Not a biggy. Here's a very good article that I think you'll enjoy:
http://autospeed.com.au/cms/title_Brake-Specific-Fuel-Consumption/A_110216/article.html

 

[fishredeemer]

For a typical car that weighs 3500 lbs, rolling resistance accounts for about 11-12 hp and aero drag the remaining 13-14 hp for a total of about 25 hp to maintain 70 mph on a dead flat road once you get up to speed. If you want to check a different speed, the rolling resistance is close to linear (30% increase = 30% more hp required) while aero drag changes as the cube of the change (30% increase = 1.3 x 1.3 x 1.3 = 2.2 times the hp, 30% decrease = 0.7 x 0.7 x 0.7 = 0.34 times the hp)

A better way to do this is to have an engine that is bigger and has a wider efficient rpm range. Ideally, you could build a drive train that can make the 25 hp needed, when running at 2000 rpm and 80-85% load, while being able to provide the maximum reserve power desired at higher rpm. Turbocharging can help fill in the hp demand as well and works very well with a diesel.

Very much like what VW is doing with their TDi Jettas and such; might be cheaper just to buy one and enjoy the fuel mileage without the work.

Don't forget that running your big engine will consume about 1/2 gallon of gas per hour at an idle, which is about the amount needed to produce 4-5 hp with the small engine, plus you're carrying around the extra weight of the normal drive train, etc, so it'll never be as efficient as a dedicated single engine system.

Most modern cars run their engines quite slow to get them into the high load efficient BSFC range and already get pretty good fuel economy for their size. My wife's full-size Buick runs at 1600 rpm on the highway and gets over 25 mpg and has more than enough reserve power. The fuel economy could be improved by up to 40% by reducing the engine size to about 1/3 the present size but would make the car all but useless on a typical drive.

Your quoted article is pretty close but the author isn't quite right on some of the reasons for the effects. Not a biggy. Here's a very good article that I think you'll enjoy:
http://autospeed.com.au/cms/title_Brake-Specific-Fuel-Consumption/A_110216/article.html

Thanks just what i need I got the rest covered

 

[Bob S]

For a 1.5 tonne car with 0.01 rolling resistance coefficient (RRC) tires, the power for rolling resistance required at 100 km/hr (62 mph or 27.7 m/s) is

P = 1500 x 9.81 x 0.01 x 27.7 = 4076 watts = 5.46 HP

The air drag is bigger, see https://www.physicsforums.com/showthread.php?t=305119&highlight=miles+gallon

[tex]P=\frac{1}{2}\rho AC_dv^3 =0.5 \cdot 1.2 \cdot 2.5 \cdot 0.35\cdot 27.7^3 = 11,160 watts = 15 HP[/tex]

where air dens = 1.2 kg per m³
frontal area of car = 2.5 m²
drag coeff = 0.35
velocity = 27.7 m/s

So total HP needed at 100 km/hr = 5.5 + 15 = 20.5 HP (at wheels)

Torque needed at 5000 RPM = [tex]T = \frac{watts \cdot 60}{2 \pi RPM}=\frac{60\cdot20.5\cdot746}{2 \pi \cdot 5000}= 29.2 Newton-meters[/tex]

[added] Torque at 3500 RPM = 41.7 N-m

Bob S

 

[jack action]

You have to take into account power losses through the transmission. You'll need a good 30-35 hp to maintain 70 mph. Use http://hpwizard.com/car-performance.html" .

 

[Bob S]

You have to take into account power losses through the transmission. You'll need a good 30-35 hp to maintain 70 mph. Use http://hpwizard.com/car-performance.html" .

Your calculator is for acceleration, not steady speed.
A good manual transmission is ~96% efficient, while an automatic is only ~85% efficient. (Did you ever wonder why manual transmissions don't need to be cooled?)
Using my formulas at 70 mph = 31.3 m/s:

Tire rolling resistance RRC = 0.01

P = m·g·RRC·v = 1500 x 9.81 x 0.01 x 31.3 = 4606 watts = 6.2 HP

Air drag at 70 mph

[tex]
P=\frac{1}{2}\rho AC_dv^3 =0.5 \cdot 1.2 \cdot 2.5 \cdot 0.35\cdot 27.7^3=16,100\;watts= 21.6\; HP [/tex]

So total HP at 70 mph is 21.6 + 6.2 = 28 HP at the wheels.

Bob S

 

[fishredeemer]

Your calculator is for acceleration, not steady speed.
A good manual transmission is ~96% efficient, while an automatic is only ~85% efficient. (Did you ever wonder why manual transmissions don't need to be cooled?)
Using my formulas at 70 mph = 31.3 m/s:

Tire rolling resistance RRC = 0.01

P = m·g·RRC·v = 1500 x 9.81 x 0.01 x 31.3 = 4606 watts = 6.2 HP

Air drag at 70 mph

[tex]
P=\frac{1}{2}\rho AC_dv^3 =0.5 \cdot 1.2 \cdot 2.5 \cdot 0.35\cdot 27.7^3=16,100\;watts= 21.6\; HP [/tex]

So total HP at 70 mph is 21.6 + 6.2 = 28 HP at the wheels.

Bob S

Thanks bob I will be back one day for some help on gear ratios
to get the rpm,s 0f the mini engine correct.
And thanks for all the imput of everyone else.
One day I will have correct data to prove the overall increase of mpg
that I believe will be obtained.
One more thing that's in mind also is propane mixer
to add power and increase fuel mpg. I have had a 6.5 diesel 1 ton
before with a bully dog propane mixer added.
Off idle operation only ofcourse but with like a 15 percent to 85 ratio
You can get considerable mpg increase ,easyly add 25 percent horse power increase
and aa cleaner burning system with lower egt,s

 

[mender]

I checked my numbers by approaching the problem from the other direction, using empirical data and and estimate of the BSFC expected from a typical production engine at 50% load; the car uses 16.8 lbs of fuel per hour at 70 mph and with a BSFC of about 0.55 lbs/hp/hr, that is around 25 hp at the road.

I used 0.015 RRC and 3700 lbs with driver to calculate the road hp at a bit over 10 hp, rounding up to 11-12 for bearing drag. Quoted specs that I just found for the car list a Cd 0f 0.31 with an approximate frontal area of 25 ft^2, giving a calculated 17 hp for aero drag; I missed that by a couple.

Using those numbers and converting to crank hp gives a BSFC of 0.487 lbs/hp/hr, which is a bit optimistic for the conditions in my opinion.

 

[xxChrisxx]

Get rid of the V8 and get an engine that's more suitably sized.
Problem solved.

If you buy a car with a 5.x L V8 you can't really be upset that it drinks fuel at a wallet sapping rate.



The power requirements are academic, they aren't the real problem, fitting a second engine and getting it to work as intended with the first engine is both complicated and expensive.

 

[mender]

Some of us want it all!

 

[xxChrisxx]

That's what turbos are for. :D :D :D

 

[mender]

fitting a second engine and getting it to work as intended with the first engine is both complicated and expensive.

And the cost will take a long time to amortize.

 

[mender]

That's what turbos are for. :D :D :D

Exactly!

 

[jack action]

Your calculator is for acceleration, not steady speed.

Once the acceleration reach zero, you are at constant speed. So in the calculator, if you enter 3500 lbm & 32 hp (engine power with a transmission) you get a top speed of 70 mph.

Tire rolling resistance RRC = 0.01

(...)

[tex]
P=\frac{1}{2}\rho AC_dv^3 =0.5 \cdot 1.2 \cdot 2.5 \cdot 0.35\cdot 27.7^3=16,100\;watts= 21.6\; HP [/tex]

So total HP at 70 mph is 21.6 + 6.2 = 28 HP at the wheels.

In the simplified version, the simulator assumes RRC = 0.013 and C_dA = 0.67 m² (you assume 0.875 m²). Who's right, who's wrong? It doesn't really matter at this point, as we have no data on the vehicle type in question.

Comparing your 28 whp (which is 29 hp at the crankshaft with a 96% eff and 33 hp with 85%) seems to be dead on with my 32 hp. If you enter 28 whp in the simulator, you get a top speed of 71 mph.

There is no surprise here, since the simulator uses the same equations as you do.

 

[AlephZero]

Who's right, who's wrong? It doesn't really matter at this point, as we have no data on the vehicle type in question.

So the intelligent thing to do would be stop putting random numbers into equations and measure it.

Find a level stretch of quiet road, accelerate to 70, knock the car out of gear, time how long it takes to decelerate to 65 or 60. Take the average time driving both ways, to cancel out the effect of small gradients, wind, etc.

You know the mass of the car, so you can find the rate of kinetic energy loss

If you can't handle that much "practical engineering", you are probably not going to succeed on the whole project anyway.

 

[fishredeemer]

Get rid of the V8 and get an engine that's more suitably sized.
Problem solved.

If you buy a car with a 5.x L V8 you can't really be upset that it drinks fuel at a wallet sapping rate.



The power requirements are academic, they aren't the real problem, fitting a second engine and getting it to work as intended with the first engine is both complicated and expensive.

Your not understanding the main concept of this idea.
Idea is mainly for increased fuel mpg. The concept is really very simple.
A big engine is necssasary for accelaration ,Climbing hills ,any type of increased load.
The idea is to aid in mpg increase ,the concept is to reduce waist of big gaseline
engine consumption by aiding with very small engine to operate at its maximum
eficententsy to obtain least amount of waisted fuel. That is a very simple concept to understand. There is an amount of fuel that is unburned during combustion plus you have the amount of fuel to create combustion that can only increase per cylinder volume size.
Then you have the increase of load for each combustion of cylinder volume. Thats 3 things
that increase per increase of engine size . The thing with automobiles is very rarely are they designed to reach maximum fuel efficentsy .
Extereme amount of varables in the powertrain, to come up with the idea combination of powertrain components and engine performance to obtain maximum iffientsy design and performance tuning. You would have to obtain all terrain ,driving habits of the driver. But the accetual reality is maximum effentintesy. The hp of mini engine would make be most effective . close to the required hp to obtain max effientsy of mpg . Being is that a bigger engine consumes a lot more fuel to apply the amount of power required to do the same job that small engine does . To the point of getting smaller car that's not the answer to increase fuel mileage of a automobile of what ever you drive, that's getting a smaller car. The idea is to lesson the amount of fuel used to
move a automobile at a selected cruise speed ,the desire for adding more aeffient smaller engine was to be able to get better fuel mileage without compromising size, weight,and performance. Sure you can do many things to increase fuel mileage to the original size engine in automobile. But then you would then be back to little engine verses big engine
to propel and automobile at the same resistance to obtain maximum fuel mileage ,the little engine will win every time . The big engine will do everything the smaller engine is not capabile of doing The smaller engine is only there to obtain maximum mpg . its not to add power are add fuel consumption . Thats as much I am going to type on that

I will correct you on the problem . The question was to obtain maxiimum fuel mileage not
figure in cost and difficulty. Engineers are every day spending more money on designing a engine just to increase efientsy of a combustible engine . Better mpg . less waist of fuel at all cost are difficulty within the maximum available resources.


Now for my expense and complication to adress.
Cost I have allready obtained a retired golf course mower for 0 dollars with low engine hrs.
But used diesel engines of this size can be had for 500 too 900 dollars without any trouble.
Difficulty very little.
drive shaft must have a pulley are gear added .No big deal . Modifiying drive shafts is simple to have done at machine shop at little cost.
pulleys of selected diameter and ratio to acquire maximum effientsy
of small engine at desired speed you will decide . A governer to control engine
speed assist on and off . Electric ingaging clutch with pulley mounted too crankshaft.
Now just mount the engine where neccasary , have driveshaft built to
connect output of engine to drive shaft.
Correct size and type of belt you selected for operation .
Cooling system would be supply from big engine radiator.
Battery supply ,and start switch hooked up.
Fuel tank of your size sellected and mounted.
Then all you have is testing and dialing in the
required injection pump throttle position to acquire your maximum fuel effentisy.
That will be the most difficult thing too accomplish of this installment.
Now you have a mini -max bi fuel automobile .
If you add propane then it would be a tri- fuel mini-max automobile.
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 .
And my cost would be very little beings I allready have golf mower with engine,
elctric clutch and driveshaft that only needs to be modified
for engine to pulley connection.Plus a fuel tank.
Being that I am a automotive tech for 30 years and can fabricate fairly well
Im not going to purchase a smaller economic auto that's not going to do the job
I need plus get better fuel mileage.
And my original question was for help figuring estimated amount of hp. to obtain sellect speed of a 3500 lb car of fair aerodynamics.
I know there's a lot of variable , specs,and data to consider.
This has been very interesting discussion with all who tune in.
Thanks again and this is a project I plan on doing but not anytime soon.
When I make this a reallity I will defintely share picts ,all the real pluses and minuses
of this project plus true fuel mileage recorded .

 

[mender]

So the intelligent thing to do would be stop putting random numbers into equations and measure it.

Find a level stretch of quiet road, accelerate to 70, knock the car out of gear, time how long it takes to decelerate to 65 or 60. Take the average time driving both ways, to cancel out the effect of small gradients, wind, etc.

You know the mass of the car, so you can find the rate of kinetic energy loss.

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.

As a guess, I'd say your truck will require at the least 30 hp (more like 35 hp) at the same speed based on your present fuel consumption, so you'll need to find a second engine.

 

[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
Theres 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.