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Old 07-29-2008, 05:23 AM   #21
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Pardon the tangent for a moment...

Originally Posted by R.I.D.E. View Post
Assuming your car is air cooled, low speed high load (sometimes called lugging)
Sometimes called lugging, but in modern fuel injected vehicles, that term is usually incorrect. Perhaps not incorrect in such an old vehicle as the OP's 1979 model. Either way it sounds like one should be careful when trying new strategies on an air-cooled engine.

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Old 07-29-2008, 07:15 AM   #22
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Originally Posted by Loserkidwac View Post
Any thoughs one accelerate at the point where HP and TQ cross over? Correct me if I'm wrong don't TQ and Hp always cross over at the same point on a dyno graph around 5200rpms? Wouldn't this be the more efficient even though HP is falling the TQ is still climbing...just a thought on my part...please discuss I'd love to here from some knowledgeably people!
They always cross at the point for a reason. You can't measure horsepower because it is a mathematical calculation of torque versus engine speed. Large V8 engines like the Vortec 8100 won't even rev to the 5250 cross over rpm. Diesels don't either. On the opposite side, the Civic Si revs way past it. That's why they can make 139 pounds of torque and make close to 200 hp. It's a numbers game.

Torque is what you want to look at, lets take my car for example. The solid line is normal engine output the others are acetone testing results.

I have two peaks to work with in my engine's power band (the first one is the intake and exhaust manifold tuned rpm and the second peak is the cams, no fancy cam switching on this 4 banger). I usually stick with having the engine rev up to 2900 and drop to 2100 at the next gear. But, if I need a little more power I can have the engine hover around 4200-5500. The engine makes over 110 pounds of rear wheel torque anywhere between 2500 and 6000 so I have pretty much the entire rev range to work with if I need it.

Brings up another question though. In my particular case, with two peaks, which one is more efficient? My logic is to get to the highest efficiency before a drop and stay there. Since it starts to fall at 3000 before it goes up again at 3750 it makes sense to me to avoid that rpm range.

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Old 07-29-2008, 09:30 AM   #23
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Personally, I still fail to see the connection between WOT torque curve characteristics and low-throttle efficiency.

There is an "internet rumor" (never substantiated, as far as I know) that your best gas mileage occurs at the same RPM as your WOT torque peak. If anyone can provide substantiation (logic, quoted sources, etc.), I'd be intrigued to know about them.

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Old 07-29-2008, 09:43 AM   #24
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The 5252 constant is for units of ft*lb of torque and horsepower. Newton meters and kilowatts use a different constant and will have a different value when the two numbers happen to be equal.
Originally Posted by dkjones96 View Post
Brings up another question though. In my particular case, with two peaks, which one is more efficient?
Maybe both, maybe neither. These charts don't show what is needed to make that determination. What is needed is a fuel consumption chart showing grams of fuel consumed, power produced, engine rpm, and at several different percentages of load. Then add gearing charts to indicate when shift points should be to minimize fuel use.
Interesting to know, but impractical to put in practice.

Here's a link to a thread discussing the BSFC for the 1Z engine in my white wagon. The BMEP is cylinder pressure, think of it as the amount of torque being produced. Also note that half the torque requires more than half the fuel. One more comment: Constant speed operation does not require maximum torque. 1900 rpm and 150 lb*ft at full load, makes nearly 55 hp. 1900 rpm in 5th is 55 mph. Steady 55 mph in 5th requires about 11 hp or 20% of load.

Have you looked at the chart and digested it a bit?
At full load (14 bar pressure, making about 150 lb*ft of torque) at 1900 rpm the fuel needed is right around 200 grams per kilowatt each hour. 200 grams times 41 kW is a 8200 grams per hour consumption rate.
Requiring less power, say only 1/5 of full load, means the pressure is 2.8 bar and torque is 30 lb*ft. At 1900 rpm that's 8.2 kW. Notice the fuel use is less efficient per kW at this load with 280 grams required per kW. Nearly 40% more fuel per kW is needed, but since the number of kW is 80% less the result is a 2300 grams per hour consumption rate.
Full load is more efficient, but the lower load produces more distance per volume of fuel.
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Old 07-29-2008, 01:31 PM   #25
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torque x rpm divided by 5252 = h.p. torque will always be more than horsepower below 5252 rpm, the same at 5252 rpm, and lower over 5252. maximum torque has less to do with optimum fuel mileage than gearing does. for example, my bike makes maximum torque at 4700 rpm, and runs 55mph at 2500 rpm. , 4700 rpm would guzzle fuel. however, if you were pulling a heavy load with a truck that makes maximum torque at 1800 rpm, and gearing allows you to operate at that speed, optimum fuel mileage under load might be at 1800 rpm. the gearing is what made it possible. if however, I was able to gear my motorcycle to operate at 4700 rpm, gas mileage would still suffer. maximum torque does not always guarantee optimum fuel mileage.

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