So what's the consensus on wide open throttle for gas mileage?

[MrSpock53]

Originally posted by: 91TTZ

Originally posted by: MrSpock53

Originally posted by: 91TTZ

Originally posted by: MrSpock53
http://www.expha.com/exphabeta...ficfuelconsumption.pdf

Above is a performance map which shows an engine's BSFC over the entire load (BMEP) and speed (N) range. A dynamometer is used to calculate brake torque and fuel flow is entered into the computation to calculate brake specific fuel consumption. This map only incorporates the wide-open throttle condition, but adding a third independent axis of throttle position can create a 3D performance map; it is noted that throttling directly affects efficiency via an inverse proportion, and maximum efficiency and minimum BSFC is realized at 100% throttle.

As ZV pointed out earlier, there's a difference in the type of efficiency we're talking about. While a 100% WOT race down the road may convert gasoline to power in the most efficient manner, that's completely different than trying to drive that same distance using the least gasoline possible. One is max BSFC and the other is max MPG.

I still don't see how WOT acceleration would negatively affect the MPG though. The idea is to minimize the pumping losses to maximize the efficiency of acceleration. Once up to speed, you would shift into the lowest gear possible, again to minimize the pumping losses. I can't think of any real reason to run at partial throttle other than to maintain a given speed.

Besides, Zenmervolt seems to pretty much agree:

Originally posted by: Zenmervolt

They all do. But you are accelerating for longer, which negates the fuel savings. For me, the difference between light acceleration and heavy acceleration is about 3 mpg difference, but with heavy acceleration, even though mpg is worse during acceleration, the overall average is better because I am accelerating for less total time.

For overall mileage, the best is to use low RPM and a large throttle opening.

ZV

Even though the engine is more efficiently converting gasoline into power during that time, you get diminishing returns in acceleration the more power you produce. For instance, if you have a car that runs 15's with 200 hp, that same car will not runs 7.5's with 400 hp. Even though you burned twice as much gas you didn't get twice the acceleration.

If you calculated the time it takes to get to 60 mph, the car with 400hp at WOT will get to that point faster than the car with 200 hp, but it won't do it twice as fast. But it will burn about twice the gas during that acceleration.

Keep in mind that the time you spend accelerating is very small compared to the time spent cruising, so the overall affect in mpg will be minimal unless you're driving in the city.

Your first statement is true. A vehicle with a 400hp engine won't get half the time of the same vehicle with the 200hp engine. This is because the amount of energy required to increase the speed of the vehicle goes up exponentially, so if one car were to finish at 120mph and the other 60mph, the car finishing at 120mph did 4 times as much work, all else being equal. While the acceleration does taper off, this does not mean that acceleration is inherently inefficient, just that physics is a bitch

I disagree with your second statement though. I believe if you took the same car with "identical" engines that ran at 200hp and 400hp, the 400hp engine would get to 60mph in half the time of the 200hp one. In order to better see why I believe this, think of it in terms of work and energy. The amount of kinetic energy required for these cars to get to 60mph is a constant. It's a specific amount of energy that does not vary with the rate of acceleration. In order to generate this kinetic energy, we use the engines to do work. Horsepower is simply a measure of how fast engines can do this work relative to each other. So if we have a specific amount of work to do (get to 60mph), the engine that can produce twice as much power will be able to do the same amount of work in half the time.

Diminishing acceleration as velocity increases is not relevant to fuel economy either way. Fuel economy is just the distance traveled over the fuel used due to our energy losses. Since we've already established that WOT at or below torque peak is the most efficient way to generate the energy necessary to get up to speed, if we need to get to 50mph in traffic the best way to do this with the WOT + shortshifting technique. Ultimately the difference in efficiency between this and light acceleration should be rather small, since acceleration is a very small portion of typical driving, for the most part we cruise. I can see how the WOT technique could be a disadvantage in some instances, for example if we are accelerating at WOT to go through a green light that ends up turning red before we can get through. In that case, although we produced the energy for acceleration relatively quick and efficiently, in the end it all got dumped into the brakes as heat. Accelerating lightly in the same situation, we would end up wasting less energy as heat. I guess the true answer on which technique is better is whether or not the efficiency gained from WOT acceleration evens out with the energy loss of having to stop at higher speeds on average.

 

[Assimilator1]

Originally posted by: Howard
I'm pretty sure a 10 000 lb car would have a lower top speed than a 1000 lb car given the same outer dimensions and power, even if for only higher bearing losses.

Yea with such a dramatic jump in weight it would ,but we weren't refering to that bigger a jump hence I ignored those factors .

91TTZ
I did mention best efficiency at peak torque.

ZV
Re fuel energy values, yep looks like I totally overlooked that! , it was a long time ago I was originally taught all this stuff .

The greater thermal efficiency is not as large a player as the lack of pumping losses.

As I said, link me to show that I'm wrong, as far as I'm concerned you are wrong unless you can prove otherwise.

Originally posted by: Zenmervolt

Originally posted by: Assimilator1
Zenmervolt
You said that even though that is the case it meant you were acclerating for less time & overall ended up with a higher MPG, by 3 ,was that calculated with the fuel computer?

No, I said that when accelerating slowly, the mpg during acceleration is approximately 3 mpg higher than the mpg during fast acceleration. I made no claim as to the amount of improvement in overall mpg caused by fast acceleration taking less time.

For example, WOT acceleration from a stop is ~5 mpg. Accelerating slowly is ~8 mpg. However, it takes 3-4 times as long to reach cruise speed by accelerating slowly. So instead of 6 seconds at 5 mpg followed by 30 seconds at 35 mpg by accelerating at WOT, a person would instead have 18-24 seconds at 8 mpg followed by 18 or 12 seconds at 35 mpg. Overall, even though the first instance has "worse" mileage during acceleration, the overall mileage for the entire 36 second cycle is superior.

Note that this is a hypothetical example to show how this would work.

ZV

Doh!, yea I mis-quoted you but I knew what you meant.

I've thought about this nr WOT practice (you should call it nr WOT not just WOT ) driving to & from work particularly & I realise I just cannot do it, traffic is often heavy & very slow so I would end up catching up the other drivers sooner & having to brake more which would be more wasteful.
As for more open conditions I realised I'd been sort of doing nr WOT for donkey years anyway! ,I guess I took more notice of Autocar than I'd thought.
Btw I say sort of as I tend not to open the throttle over 50% in 1st or 2nd as it makes gearchanging very rough, the clutch on my BMW is rather snatchy .

 

[91TTZ]

Originally posted by: MrSpock53

I disagree with your second statement though. I believe if you took the same car with "identical" engines that ran at 200hp and 400hp, the 400hp engine would get to 60mph in half the time of the 200hp one.

This is completely, entirely, 100% incorrect.

Think about it:

Take an old Honda Civic with 100 hp. It does 0-60 in about 8.6 seconds. You're telling me that if you gave it 200 hp, it would do 0-60 in 4.3 seconds? 400 hp would mean 2.15 seconds?

Or take the quarter mile which would give a better example of acceleration. With 100 hp it does it in roughly 16 seconds. You're telling me that with 200 hp it's going to do it in 8 seconds? 400 hp it's doing it in 4 seconds?

To put it into perspective, a top fuel dragster does the quarter mile in about 4 seconds, it's hundreds of pounds lighter than the Civic in my example, and it has 7,000 hp.

 

[91TTZ]

To add to my above comment, since we already know that acceleration does not scale linearly with HP while fuel consumption does scale linearly with HP, it's easy to figure out that you're going to get diminishing returns the faster you want to accelerate.

Let's say you want to accelerate to 60 mph. In our Civic example, with 100 hp it will get to 60 in about 8.6 seconds. That means that it's consuming 100 hp worth of fuel for 8.6 seconds before lightly cruising to maintain your speed. If you wanted to get to 60 in half the time, you're going to need way more than twice the horsepower. That means you'll be burning fuel at a rate much higher than 2x for half the time.

If you want to see the math involved in figuring out how much HP it would take you to accelerate at a certain rate, look here: link or here: Text

If you want to see the math involved in figuring out how much fuel you need to burn to produce x amount of horsepower, you can get a rough idea here: at the bottomlink



Let's use my car as an example. I've already dyno'd it so I already know that I have 339 HP and 389 lbs of torque at the rear wheels. I already know that my car weighs about 3500 lbs, and with me in it it's about 3700 lbs. I already ran a 13.0 in the quarter, so let's see how these figures add up. The HP calculator says that a 3700 lb car will need about 333 hp at the wheels to do the quarter mile in 13.0 seconds. In order to do it in 6.5 seconds I would need 2,663 Hp at the wheels.

Now, for the final calculation using the automotive math I linked to:

If you want to bring your 3000, 200 hp car up to 60 mph at WOT, you're going to get to 60 mph in about 6.9 seconds and during that time you'll be burning fuel at the rate of 16.67 gallons per hour. On the other hand if you give it less gas and only produce 100 hp, you'll get to 0-60 in 11.6 seconds and burn fuel at a rate of 8.33 gallons per hour.

Cost for your 200 hp burst to 60 mph: .032 gallons
Cost for your 100 hp burst to 60 mph: .027 gallons

 

[Zenmervolt]

Originally posted by: 91TTZ
To add to my above comment, since we already know that acceleration does not scale linearly with HP while fuel consumption does scale linearly with HP, it's easy to figure out that you're going to get diminishing returns the faster you want to accelerate.

Invalid comparison though.

Remember that power figures are always arrived at under WOT. If your car develops 100 hp at 3,000 RPM, then that's at WOT at 3,000 RPM. My 951 makes about 275 hp at the rear wheels at around 4,500 RPM at WOT, but if I'm cruising down the freeway at a steady 70 mph with the engine spinning at 4,500 RPM and not accelerating, then the engine is only making around 35 hp. All of the remaining potential power is being choked off by the throttle butterfly.

Your comparisons are assuming that an engine operates at a fixed efficiency throughout the RPM range. The efficiency of an engine is vastly improved at WOT because the pumping losses essentially vanish and the horsepower is produced much more efficiently. All of the equations you cite for listing the fuel required to make hp assume that the engine is operating at WOT.

Using some rough numbers for WOT acceleration at 60 mph versus steady-state 60 mph in my 951:

5 mpg 275 hp = 55 hp/mpg
27 mpg 35 hp = 1.30 hp/mpg

5 mpg at 60 mph = 12 gallons/hour
27 mpg at 60 mph = 2.22 gallons/hour

12 gallons/hour at 275 hp = 22.92 hp per gallon per hour
2.22 gallons/hour at 35 hp = 15.77 hp per gallon per hour

You can clearly see that the engine is over 45% more efficient at producing power when at WOT. compared to operating with a restrictive throttle. At WOT, the engine makes significantly more power per unit of fuel used.

ZV

 

[91TTZ]

Originally posted by: Zenmervolt

Invalid comparison though.

Remember that power figures are always arrived at under WOT. If your car develops 100 hp at 3,000 RPM, then that's at WOT at 3,000 RPM. My 951 makes about 275 hp at the rear wheels at around 4,500 RPM at WOT, but if I'm cruising down the freeway at a steady 70 mph with the engine spinning at 4,500 RPM and not accelerating, then the engine is only making around 35 hp. All of the remaining potential power is being choked off by the throttle butterfly.

Your comparisons are assuming that an engine operates at a fixed efficiency throughout the RPM range. The efficiency of an engine is vastly improved at WOT because the pumping losses essentially vanish and the horsepower is produced much more efficiently. All of the equations you cite for listing the fuel required to make hp assume that the engine is operating at WOT.

Using some rough numbers for WOT acceleration at 60 mph versus steady-state 60 mph in my 951:

5 mpg 275 hp = 55 hp/mpg
27 mpg 35 hp = 1.30 hp/mpg

5 mpg at 60 mph = 12 gallons/hour
27 mpg at 60 mph = 2.22 gallons/hour

12 gallons/hour at 275 hp = 22.92 hp per gallon per hour
2.22 gallons/hour at 35 hp = 15.77 hp per gallon per hour

You can clearly see that the engine is over 45% more efficient at producing power when at WOT. compared to operating with a restrictive throttle. At WOT, the engine makes significantly more power per unit of fuel used.

ZV

While that is interesting, how can I be sure that your first couple of figures are correct? Since all the math is based on them I'd like to be sure that they're accurate.

You're getting 1/5.4th the gas mileage but getting 7.86x the power. Something doesn't seem right there. Maybe it's because with the steady state cruising number, all the power is going towards fighting wind resistance while with the WOT acceleration, some is going towards fighting wind resistance while the surplus is going towards accelerating the vehicle?

Ideally I'd like to take mpg out of the equation and compare HP vs. fuel consumption instead.

 

[Zenmervolt]

Originally posted by: 91TTZ

Originally posted by: Zenmervolt

Invalid comparison though.

Remember that power figures are always arrived at under WOT. If your car develops 100 hp at 3,000 RPM, then that's at WOT at 3,000 RPM. My 951 makes about 275 hp at the rear wheels at around 4,500 RPM at WOT, but if I'm cruising down the freeway at a steady 70 mph with the engine spinning at 4,500 RPM and not accelerating, then the engine is only making around 35 hp. All of the remaining potential power is being choked off by the throttle butterfly.

Your comparisons are assuming that an engine operates at a fixed efficiency throughout the RPM range. The efficiency of an engine is vastly improved at WOT because the pumping losses essentially vanish and the horsepower is produced much more efficiently. All of the equations you cite for listing the fuel required to make hp assume that the engine is operating at WOT.

Using some rough numbers for WOT acceleration at 60 mph versus steady-state 60 mph in my 951:

5 mpg 275 hp = 55 hp/mpg
27 mpg 35 hp = 1.30 hp/mpg

5 mpg at 60 mph = 12 gallons/hour
27 mpg at 60 mph = 2.22 gallons/hour

12 gallons/hour at 275 hp = 22.92 hp per gallon per hour
2.22 gallons/hour at 35 hp = 15.77 hp per gallon per hour

You can clearly see that the engine is over 45% more efficient at producing power when at WOT. compared to operating with a restrictive throttle. At WOT, the engine makes significantly more power per unit of fuel used.

ZV

While that is interesting, how can I be sure that your first couple of figures are correct? Since all the math is based on them I'd like to be sure that they're accurate.

You're getting 1/5.4th the gas mileage but getting 7.86x the power. Something doesn't seem right there. Maybe it's because with the steady state cruising number, all the power is going towards fighting wind resistance while with the WOT acceleration, some is going towards fighting wind resistance while the surplus is going towards accelerating the vehicle?

Ideally I'd like to take mpg out of the equation and compare HP vs. fuel consumption instead.

My car has been dynoed, the 275 hp figure is accurate for 60 mph, 4th gear, and WOT.

Drag increases with the square of speed and it takes 8x the power to go twice as fast. With 275 hp (less at redline, actually), my car has a top speed of 172 mph. The power required to go only 86 mph is 275/8 (actually, something like 250/8, but I'm intentionally overstating to make my estimate conservative) or 34.375 hp. Now, this is at 86 mph, so at 60 mph, the engine is making even less than the 35 hp I use in my figures (probably closer to 25 hp), but again, I wanted to ensure that I was being conservative.

MPG under acceleration uses the assumption that my injectors are at 100% duty cycle and MPG under cruise is measured based on mileage obtained with 100% freeway driving.

You're welcome to check my math on the above, but the numbers are close enough for a "back of the envelope" calculation that will accurately illustrate the principle.

MPG is essentially out of the equation in the final equation, but we're stuck with having a time component since one of our bogeys is fuel consumption, which has to be measured over time. That is the reason why I converted from mpg to gallons/hour though.

The reason that power increases more than mileage decreases is because the engine has to work against the throttle restriction at cruising speed. Creating manifold vacuum is a tremendous waste of power, which is what the figures indicate. When comparing two engines to each other, power ratings are taken at WOT. This equalizes things and does tend to make power scale linearly (or roughly so) with fuel consumption. However, if one engine has a large efficiency gain, that will throw off the roughly linear relationship. Using less than WOT introduces an intentional inefficiency to prevent the engine from accelerating out of control when there is no load on it.

ZV

 

[JulesMaximus]

Originally posted by: Zenmervolt

Originally posted by: 91TTZ
To add to my above comment, since we already know that acceleration does not scale linearly with HP while fuel consumption does scale linearly with HP, it's easy to figure out that you're going to get diminishing returns the faster you want to accelerate.

Invalid comparison though.

Remember that power figures are always arrived at under WOT. If your car develops 100 hp at 3,000 RPM, then that's at WOT at 3,000 RPM. My 951 makes about 275 hp at the rear wheels at around 4,500 RPM at WOT, but if I'm cruising down the freeway at a steady 70 mph with the engine spinning at 4,500 RPM and not accelerating, then the engine is only making around 35 hp. All of the remaining potential power is being choked off by the throttle butterfly.

Your comparisons are assuming that an engine operates at a fixed efficiency throughout the RPM range. The efficiency of an engine is vastly improved at WOT because the pumping losses essentially vanish and the horsepower is produced much more efficiently. All of the equations you cite for listing the fuel required to make hp assume that the engine is operating at WOT.

Using some rough numbers for WOT acceleration at 60 mph versus steady-state 60 mph in my 951:

5 mpg 275 hp = 55 hp/mpg
27 mpg 35 hp = 1.30 hp/mpg

5 mpg at 60 mph = 12 gallons/hour
27 mpg at 60 mph = 2.22 gallons/hour

12 gallons/hour at 275 hp = 22.92 hp per gallon per hour
2.22 gallons/hour at 35 hp = 15.77 hp per gallon per hour

You can clearly see that the engine is over 45% more efficient at producing power when at WOT. compared to operating with a restrictive throttle. At WOT, the engine makes significantly more power per unit of fuel used.

ZV

Yeah, that's good in theory but in the real world it's pretty much impossible or impractical not to mention irresponsible to drive like this.

 

[91TTZ]

Originally posted by: Zenmervolt


Drag increases with the square of speed and it takes 8x the power to go twice as fast. With 275 hp (less at redline, actually), my car has a top speed of 172 mph. The power required to go only 86 mph is 275/8 (actually, something like 250/8, but I'm intentionally overstating to make my estimate conservative) or 34.375 hp. Now, this is at 86 mph, so at 60 mph, the engine is making even less than the 35 hp I use in my figures (probably closer to 25 hp), but again, I wanted to ensure that I was being conservative.




ZV

Wouldn't it 4x the power to go twice as fast? If drag increases with the square of speed it wouldn't be 8x, right? Looking around on the net I've seen both quoted as being the case.

 

[Assimilator1]

I seem to recall that once the speed goes over ~150 mph the amount of power required goes up massivley.
I can't remember the name of it but if you draw a graph for power needs vs speed its a curved graph not a straight (linear) one. (Logorithmic maybe?)
Don't know if that helps much lol .

Jules
As in your 1st post you're missing the point that (excluding comparisions for full power) driving using the (near) WOT practice the engine is not revved up to the redline, but at the most roughly 3000 RPM.
This can be practical & is not irresponsible in itself. Naturally you can't even use 'low rev' nr WOT in stop start traffic .
But anyway the idea is that short low rev bursts of nr WOT (using the engine in it's most efficient mode) will improve fuel economy as long as you don't have to slow down immediatley or soon after .

Originally posted by: Zenmervolt

Originally posted by: 91TTZ
To add to my above comment, since we already know that acceleration does not scale linearly with HP while fuel consumption does scale linearly with HP, it's easy to figure out that you're going to get diminishing returns the faster you want to accelerate.

Invalid comparison though.

Remember that power figures are always arrived at under WOT. If your car develops 100 hp at 3,000 RPM, then that's at WOT at 3,000 RPM. My 951 makes about 275 hp at the rear wheels at around 4,500 RPM at WOT, but if I'm cruising down the freeway at a steady 70 mph with the engine spinning at 4,500 RPM and not accelerating, then the engine is only making around 35 hp. All of the remaining potential power is being choked off by the throttle butterfly.

ZV

..... & by the amount of fuel metered.

The throttle plate is only needed because petrol engines can't handle any significant changes in fuel/air mixture ratio ,especially with a cat. Otherwise you could adjust power just by adjusting the amount of fuel delivered ,like a Diesel. As you know .

 

[Zenmervolt]

Originally posted by: 91TTZ

Originally posted by: Zenmervolt


Drag increases with the square of speed and it takes 8x the power to go twice as fast. With 275 hp (less at redline, actually), my car has a top speed of 172 mph. The power required to go only 86 mph is 275/8 (actually, something like 250/8, but I'm intentionally overstating to make my estimate conservative) or 34.375 hp. Now, this is at 86 mph, so at 60 mph, the engine is making even less than the 35 hp I use in my figures (probably closer to 25 hp), but again, I wanted to ensure that I was being conservative.




ZV

Wouldn't it 4x the power to go twice as fast? If drag increases with the square of speed it wouldn't be 8x.

Nope, 8x the power. 4x the drag multiplied by 2x the velocity is 8x the total power. You have to double speed as well as deal with quadruple the drag, so the two factors are multiplied. It took me a while to wrap my head around it, but it does make sense.

To overcome drag, 4x the work is being done to maintain velocity. However, since the speed has doubled, the work has half the time to get done. 4x the work done in 1/2 the amount of time is 8x the power.

ZV

 

[91TTZ]

Originally posted by: Zenmervolt

Nope, 8x the power. 4x the drag multiplied by 2x the velocity is 8x the total power. You have to double speed as well as deal with quadruple the drag, so the two factors are multiplied. It took me a while to wrap my head around it, but it does make sense.

To overcome drag, 4x the work is being done to maintain velocity. However, since the speed has doubled, the work has half the time to get done. 4x the work done in 1/2 the amount of time is 8x the power.

ZV

It would seem like as long as the car's engine had more power than is required to overcome the drag at x speed, its speed would continue to creep up until it reached a point of equilibrium.


For instance, let's say drag consumes 10 hp at 20 mph and 40 hp at 40 mph. It would seem that if you had 40 hp and hit the gas while going 10 mph, your speed would continue to increase until you reached the point of equilibrium at 40 mph where your engine can no longer overcome the force of drag. It wouldn't seem right that you'd need 80 hp to double your speed.

 

[Zenmervolt]

Originally posted by: 91TTZ

Originally posted by: Zenmervolt

Nope, 8x the power. 4x the drag multiplied by 2x the velocity is 8x the total power. You have to double speed as well as deal with quadruple the drag, so the two factors are multiplied. It took me a while to wrap my head around it, but it does make sense.

To overcome drag, 4x the work is being done to maintain velocity. However, since the speed has doubled, the work has half the time to get done. 4x the work done in 1/2 the amount of time is 8x the power.

ZV

It would seem like as long as the car's engine had more power than is required to overcome the drag at x speed, its speed would continue to creep up until it reached a point of equilibrium.


For instance, let's say drag consumes 10 hp at 20 mph and 40 hp at 40 mph. It would seem that if you had 40 hp and hit the gas while going 10 mph, your speed would continue to increase until you reached the point of equilibrium at 40 mph where your engine can no longer overcome the force of drag. It wouldn't seem right that you'd need 80 hp to double your speed.

You're still only accounting for the drag and missing the fact that you have to spin the driveshaft twice as fast at 40 mph as you do at 20 mph. That means the work has to be done twice as fast, which requires twice the power. So 4x the power to overcome the drag multiplied by 2x the rate at which work is being done. 8x the overall power.

You have to account for the fact that even absent of all drag, it will take 2x the power to double the speed. You are neglecting this.

From the wikipedia article, which is accurate in terms of its equations (emphasis added):

Note that the power needed to push an object through a fluid increases as the cube of the velocity. A car cruising on a highway at 50 mph (80 km/h) may require only 10 horsepower (7.5 kW) to overcome air drag, but that same car at 100 mph (160 km/h) requires 80 hp (60 kW). With a doubling of speed the drag (force) quadruples per the formula. Exerting four times the force over a fixed distance produces four times as much work. At twice the speed the work (resulting in displacement over a fixed distance) is done twice as fast. Since power is the rate of doing work, four times a work in half the time requires eight times the power.

Link

ZV

 

[JulesMaximus]

Originally posted by: Assimilator1
I seem to recall that once the speed goes over ~150 mph the amount of power required goes up massivley.
I can't remember the name of it but if you draw a graph for power needs vs speed its a curved graph not a straight (linear) one. (Logorithmic maybe?)
Don't know if that helps much lol .

Jules
As in your 1st post you're missing the point that (excluding comparisions for full power) driving using the (near) WOT practice the engine is not revved up to the redline, but at the most roughly 3000 RPM.
This can be practical & is not irresponsible in itself. Naturally you can't even use 'low rev' nr WOT in stop start traffic .
But anyway the idea is that short low rev bursts of nr WOT (using the engine in it's most efficient mode) will improve fuel economy as long as you don't have to slow down immediatley or soon after .

Originally posted by: Zenmervolt

Originally posted by: 91TTZ
To add to my above comment, since we already know that acceleration does not scale linearly with HP while fuel consumption does scale linearly with HP, it's easy to figure out that you're going to get diminishing returns the faster you want to accelerate.

Invalid comparison though.

Remember that power figures are always arrived at under WOT. If your car develops 100 hp at 3,000 RPM, then that's at WOT at 3,000 RPM. My 951 makes about 275 hp at the rear wheels at around 4,500 RPM at WOT, but if I'm cruising down the freeway at a steady 70 mph with the engine spinning at 4,500 RPM and not accelerating, then the engine is only making around 35 hp. All of the remaining potential power is being choked off by the throttle butterfly.

ZV

..... & by the amount of fuel metered.

The throttle plate is only needed because petrol engines can't handle any significant changes in fuel/air mixture ratio ,especially with a cat. Otherwise you could adjust power just by adjusting the amount of fuel delivered ,like a Diesel. As you know .

Wouldn't you just bog the hell out of the engine doing that?

 

[Assimilator1]

No you don't let the revs fall that low , for instance in my car I could rev from 1500-3000 ,mine doesn't 'bog down' until about 1200-1300 RPM. Though it is probably fairly inefficient at those revs. I might even be better off staying between 2000-3500 as it doesn't pull much below 2000 ,indicating it's off the main torque band/curve. Though that's more of guess & not based on any figures I have (other than I know peak torque is at 4000 RPM).
I suppose I could measure it with the fuel computer,........ anyway you can use wide throttle openings at lowish revs without bogging it down (I think you're thinking of carburettor engines ), exactly where will depend on the torque band of the engine.

 

[91TTZ]

Originally posted by: Zenmervolt

You're still only accounting for the drag and missing the fact that you have to spin the driveshaft twice as fast at 40 mph as you do at 20 mph. That means the work has to be done twice as fast, which requires twice the power. So 4x the power to overcome the drag multiplied by 2x the rate at which work is being done. 8x the overall power.

You have to account for the fact that even absent of all drag, it will take 2x the power to double the speed. You are neglecting this.
ZV

In the absence of drag, it does not take twice the power to go twice as fast. In fact, without drag you could have a 1 hp engine and accelerate to thousands of miles per hour, it'll just take you a while to reach that speed. Without all that drag you could simply gear it up and keep accelerating since we're not limited by a time constraint. You'd slowly keep on accelerating, and the longer you stay on the gas the faster you'd get.

Now, if you were to say that it requires 8x the power to accelerate from 60-120 in the same amount of time that it takes to accelerate from 0-60, then I could believe that. But we're not talking about that, we're talking about ultimate top speed, which has no time limit. I think that it's all power vs. drag.

Go easy on me because I'm having trouble understanding this. What seems to make sense somehow isn't the case, and what is the case makes no sense to me.

 

[Zenmervolt]

Originally posted by: 91TTZ

Originally posted by: Zenmervolt

You're still only accounting for the drag and missing the fact that you have to spin the driveshaft twice as fast at 40 mph as you do at 20 mph. That means the work has to be done twice as fast, which requires twice the power. So 4x the power to overcome the drag multiplied by 2x the rate at which work is being done. 8x the overall power.

You have to account for the fact that even absent of all drag, it will take 2x the power to double the speed. You are neglecting this.
ZV

In the absence of drag, it does not take twice the power to go twice as fast. In fact, without drag you could have a 1 hp engine and accelerate to thousands of miles per hour, it'll just take you a while to reach that speed. Without all that drag you could simply gear it up and keep accelerating since we're not limited by a time constraint. You'd slowly keep on accelerating, and the longer you stay on the gas the faster you'd get.

Now, if you were to say that it requires 8x the power to accelerate from 60-120 in the same amount of time that it takes to accelerate from 0-60, then I could believe that. But we're not talking about that, we're talking about ultimate top speed, which has no time limit. I think that it's all power vs. drag.

Go easy on me because I'm having trouble understanding this. What seems to make sense somehow isn't the case, and what is the case makes no sense to me.

I think that I've realized why the hangup. When I'm talking about drag, I'm talking about aerodynamic drag. Not frictional drag like the drag from tires or the driveline. It's the aerodynamic drag that quadruples when speed doubles. I've been unclear by using the generic term "drag".

ZV

 

[91TTZ]

Originally posted by: Zenmervolt

Originally posted by: 91TTZ

Originally posted by: Zenmervolt

You're still only accounting for the drag and missing the fact that you have to spin the driveshaft twice as fast at 40 mph as you do at 20 mph. That means the work has to be done twice as fast, which requires twice the power. So 4x the power to overcome the drag multiplied by 2x the rate at which work is being done. 8x the overall power.

You have to account for the fact that even absent of all drag, it will take 2x the power to double the speed. You are neglecting this.
ZV

In the absence of drag, it does not take twice the power to go twice as fast. In fact, without drag you could have a 1 hp engine and accelerate to thousands of miles per hour, it'll just take you a while to reach that speed. Without all that drag you could simply gear it up and keep accelerating since we're not limited by a time constraint. You'd slowly keep on accelerating, and the longer you stay on the gas the faster you'd get.

Now, if you were to say that it requires 8x the power to accelerate from 60-120 in the same amount of time that it takes to accelerate from 0-60, then I could believe that. But we're not talking about that, we're talking about ultimate top speed, which has no time limit. I think that it's all power vs. drag.

Go easy on me because I'm having trouble understanding this. What seems to make sense somehow isn't the case, and what is the case makes no sense to me.

I think that I've realized why the hangup. When I'm talking about drag, I'm talking about aerodynamic drag. Not frictional drag like the drag from tires or the driveline. It's the aerodynamic drag that quadruples when speed doubles. I've been unclear by using the generic term "drag".

ZV

I understood that part, it's the part about why it goes up by 8x instead of 4x that I'm hung up about.

 

[Throckmorton]

Originally posted by: 91TTZ

Originally posted by: Zenmervolt

Originally posted by: 91TTZ

Originally posted by: Zenmervolt

You're still only accounting for the drag and missing the fact that you have to spin the driveshaft twice as fast at 40 mph as you do at 20 mph. That means the work has to be done twice as fast, which requires twice the power. So 4x the power to overcome the drag multiplied by 2x the rate at which work is being done. 8x the overall power.

You have to account for the fact that even absent of all drag, it will take 2x the power to double the speed. You are neglecting this.
ZV

In the absence of drag, it does not take twice the power to go twice as fast. In fact, without drag you could have a 1 hp engine and accelerate to thousands of miles per hour, it'll just take you a while to reach that speed. Without all that drag you could simply gear it up and keep accelerating since we're not limited by a time constraint. You'd slowly keep on accelerating, and the longer you stay on the gas the faster you'd get.

Now, if you were to say that it requires 8x the power to accelerate from 60-120 in the same amount of time that it takes to accelerate from 0-60, then I could believe that. But we're not talking about that, we're talking about ultimate top speed, which has no time limit. I think that it's all power vs. drag.

Go easy on me because I'm having trouble understanding this. What seems to make sense somehow isn't the case, and what is the case makes no sense to me.

I think that I've realized why the hangup. When I'm talking about drag, I'm talking about aerodynamic drag. Not frictional drag like the drag from tires or the driveline. It's the aerodynamic drag that quadruples when speed doubles. I've been unclear by using the generic term "drag".

ZV

I understood that part, it's the part about why it goes up by 8x instead of 4x that I'm hung up about.

Hmm... The engine should be encountering 4x the resistance, and turn twice as fast right? That would make 8x.

But I don't think that's right. I think the engine encounters 2x the resistance force, because of drag increasing, BUT also has to spin 2x as fast. That combines to get the 4x number. Is that right??