Understanding aero vs throttle pos.
 

I read a post the other day (can't seem to locate it) that confused me a bit. Maybe I misread it.

Here is what I know: Aero improvements cannot change engine rpm for a set gear and speed because gear ratios are fixed. Aero improvements can only decrease the required throttle position to maintain that speed, by putting more engine power towards forward motion (versus overcoming wind resistance). One could opt for a taller final drive ratio, but that's a whole different story.

But here's the part that tripped me up: In some situations, the lighter throttle position invites more pumping losses? I can't wrap my head around a situation in which better Cd, which means lighter cruising throttle pos., could have a negative result.

Perhaps pumping losses are increased, but it's still always a net gain - right?

I'll try to see if I can come across the post, but I think I explained it correctly.

The decreased in aerodynamic drag is a much larger factor than pumping losses. The increased pumping losses are more than offset.

Correct, it's always a net gain. I read one paper (don't remember title, when, or where) that said the real world gain for aero improvements alone is about half the total drag reduction. You get the rest of the gain by gear ratio changes.

And by gearing up you require a larger throttle opening, eliminating the additional pumping losses you originally created.

Ah! And now it's clear. Thanks for filling in the missing pieces of the puzzle - makes total sense.

An other way to think about it is that at Wide Open Throatle the pumping loses are the lowest but the fuel consumed is the greatest. Whereas at idle the pumping loses are the greatest and the fuel consumed is the lowest. In spark ignition gas engines the throatle plate controls the mass of air/fuel the combustion chamber sees... to evercome resistance; rolling, aero...

Makes sense to me.
I would also say that with better aero, you're able to coast more often and for a longer distance. This reduces your rpms to either zero or idle speed, depending on your preferred coasting method. Of course, that's once you've added coasting to your "tool kit". So depending on your route and the hills etc., I believe coasting WITH improved aero can give more benefit than just what you gain from drag reduction alone.

No it doesn't. Taller gearing means lower RPM for a given speed therefore reduces manifold vacuum (higher absolute pressure) and reduced pumping losses. The amount of power to overcome rolling resistance and aerodynamic drag remains the same.

This should mean diesel or electric powered vehicles will respond better to aerodynamic improvements than gasoline counterparts....

- Unless we use pulse & glide

Sorry - I don't see the logical path for either conclusion.
Unless the point you're making is that you need to P&G to get the maximum benefit of an aero mod, when driving a gasoline car. I think I'll agree with you on that.

My gas-powered Civic has responded very well to aero improvements and I'd be very reluctant to give them up. In fact, I'm working on a new mod almost "as we speak".

Even if you could show a logic for your double conclusion above, saying that one type would respond "better" to aero improvements would be an apples to oranges comparison.

I could argue that gasoline cars, with their inherent losses of thermal inefficiency, need every bit of help that we can bring to bear.

Well - diesel and electric don't suffer as much from pumping losses so there is no such thing as a more restricted air intake when better aerodynamics results in less throttle.

However, if you p&g with the gasoline engine, you don't increase your suffering from pumping losses because you only use the engine with open throttle.

As been already said, you will always gain with improved aerodynamics, even in steady state driving. Engine efficiency probably drops a little bit more with a gasoline engine though.

Sounds right to me.
Apparently there's plenty I could learn abut diesel - I never owned one, and drove one only briefly nearly 30 years ago (VW Rabbit).

In my diesel, coasting in neutral with the engine on, the ScanGuage reads 300 to 400 instantaneous mpg. There is no throttle plate creating vacuum and choking efficiency at low load conditions. The computer injects only enough fuel to keep the engine turning over. With better aero I can coast in gear more on long grades using no fuel at all. Fuel is cut off at anything over 900 rpm while engine braking (overrun) in the VW.

Sounds right to me. Gasoline engine efficiency drops off faster at part throttle than either diesel or electric. So those of us with gasoline engines get around that by increasing P&G as we decrease aero drag.

Power remains the same, but RPM is lower, therefore torque must be higher, requiring a larger throttle opening.

No, with taller gearing for a given road speed, you still need the same amount of air to send the same amount of torque to the wheels. RPM decreases, absolute manifold pressure increases, throttle angle remains the same, torque remains the same. On a naturally aspirated engine you reach you peak torque for a given RPM when manifold pressure is at 100kpa regardless of the throttle angle. At low RPMs you can reach peak torque at 30-50% throttle.

Power = torque X RPM. If you decrease RPM, you must increase torque to maintain a given power.

You're thinking about peak torque at any given RPM. At part throttle cruise on flat ground you don't need that. You're just playing around inside the torque curve area. If you install taller gearing your airflow and torque remain the same even though you reduced the RPM for a given cruising speed. If you have a good OBD2 tool (scan gauge, ultra gauge, Torque app etc), get on the highway and cruise at a fixed speed in 4th gear on flat ground, note the throttle angle and MAF reading, then shift to 5th. You'll see the same angle and MAF reading despite the reduced RPM. Your injector pulse width will be different because fuel pressure is vacuum referenced. Thats why all OBD2 systems use MAF not injector PW to calculate fuel consumption.

The equation holds at any RPM, torque, or power. Just like E = I X R, it's a mathematical and physical fact. If you decrease rpm for a given power, you must increase torque. If you lower RPM at the same torque, you have decreased power.

Ok if you say so but this has nothing to do with throttle angle or the amount of air consumed. Throttle angle stays the same, MAF load stays the same given a fixed speed and taller gearing. If you don't believe me, try it on your own car. Its very simple test. Anybody with an OBD2 can do this.

if you mean mpg monitors, they use MAF (or figure it out from MAP/RPM/temp) because there isn't an obd2 pid for pulsewidth, and they hope for stoic. Vacuum referenced regulator is good in that it standardizes the fuel delivered for a given amount of open injector time regardless of the manifold pressure.

If you mean OBD2 ECUs use MAF and not injector PW, well, that is because injector PW is an OUTPUT from the ECU ?!?

Bsfc
 

I believe it was Gino Sovran that published that unless gear-matching was accomplished along with drag reduction,up to 60% of the streamlining benefit could be lost,as the engine BSFC is moved into a less efficient operating regime.Hucho covers this in his book.Chrysler discovered this firsthand in 1934 when they aero-modded the DeSoto Airflow test mule.
I may have just experienced this with the trailer project.Only further,higher speed testing will tell.

Patric & tjts1
If I accept that Power = Torque x rpm [and I do...] and I accept that the throatle position remains the same at a lower rpm [and I do...] and I accept that the engine pumps the same mass of air at both higher and lower rpm's and therefore produces the same amount of power, then the force provided must be greater for each revolution ie torque is greater at the lower rpm. Which if I am not mistaken is consistent with the throatle position being the same at both higher and lower rpm.
Please correct any errors I may have made with my high school phisycs and math limited understanding.

Please explain how the engine increased torque with the same amount of air.

Also, please explain how the engine (a positive displacement pump) pulled in the same amount of air at a lower rpm with the throttle plate at the same position.

The volume [displacement] of air is the same for every cycle and at all rpm. What changes is the mass of the air in the cylinder when the intake valve closes. The mass of air is controled by the throatle plate in spark ignition engines, therefore if more torque is produced the throatle plate should have to be more open [less restiction].
To produce the same amount of power the engine must consume the same mass of of air and fuel with the same thermal efficiancey at the lower and higher rpm. So the volume of air is directly proportional to the rpm the mass is changed by the throatle plate.
The only thing that changes is the amount of time [longer] the intake valve is open at the lower rpm so with the same rate of flow [mass] past the throatle plate the longer time would allow a greater mass of air to enter the combustion chamber. The question that I don't know the answer to is; is that enough to raise the torque as much as is needed to produce the same amount of power, I some how doubt it. So unless the time difference is enough then the plate has to be less restrictive [more open] to allow the same mass of air to enter the engine at the lower rpm.

Do I have it right yet?:confused::o