ok. when I was pulling my 17 foot 3200 pound camp trailer home yesterday, I noticed that at 60 in Drive (not OD) the aero seemed to have more go power than at 45 in drive. I am wondering if the rpm range at 60 is more within the power range of the 4.0, and then wondering about MPG, if 60 in drive is better.

but then I wondered about wind resistance. the trailers face is 8 feet square. 64 square feet total. and I have a short tongue. what is the pounds of resistance pulled at 45 and 60 mph? of course the van has a certain amount of draft area. so, that 64 sq foot area is reduced by the vans area. sooooo, im stumped.

any answers??

 

One of the laws of physics is a body at rest stays at rest and abody in motion tends to stay in motion untill a force is exerted on it. Basically I think you are feeling the effects of connetic energy and your RPMs are higher at 60 so you are making more horse power at 60 than 45. Your gas mileage will be less and here is why. An internal combustion engine is not an efficiant machine. If you put a engine on a stand and run it at idle a gallon of gas will last longer than at any other speed above idle. Its simple for the engine to turn faster you need to give it more gas. Now you van has gears,well your first 3 gears are gear reduction because an engine is not strong enough to get a vehical going from a dead stop with out reducing the final drive ( some 4 speed an 5 speed manual transmitions have 4th gear at 1:1 Know as direct drive. it means for every 1RPM you engine turns your drive shaft truns at the same RPM and this is what I mean in the above). So all this boils down to your van or any vehical will get its best gas miliage when you are in the highest gear possible with the least amount of gas being put into the engine. Over Drive means that your final transmittion gear will have your engine turning slower than the drive shaft and saves you gas as your engine is working less. So you see you van with Over drive disengaged is already in third gear at 45 so to go 60 you need to put more fuel through the engine. Our vans shift into over drive around 50 so even to go 51 in overdrive you need more gas so less gas mileage but because of the lower engine speed you will not see the effect as quickly as in the lower gears.

 

Let's see if I can help here. At 60 mph, that's about 27 m/s. Air density is about 1.2kg/cubic meter. The wind resistance is 1/2*density*square of velocity*drag coefficient*area. Assuming the drag coefficient is about 0.5 for that ugly trailer , that comes out to about 220 Newtons/square meter or about 4.6 pounds per square foot. The wind drag on your entire vehicle comes to about 300 pounds, give or take 100%

At 45 mph, the velocity is only 3/4 of 60mph, so the drag will be 300 pounds times the square of 3/4, or about 170 pounds.

But of course, we're neglecting road friction. When you add that in, the actual number could be twice as much

One major reason why the gas mileage sucks at high speed is the fact that the drag varies like the square of velocity. To move a distance, the work required is the product of distance and drag. If the distance remains the same, going at 60mph requires 4 times the work required when you are going at 30mph, but you get there twice as fast . But then again, we're only talking wind resistance. When you add in friction, the effect is much less dramatic, and the transmission is supposed to alleviate some of that.

However, once you get to the point of say 60mph, wind resistance starts to dominate and your mileage will decrease very fast with speed.

 

WOW

I knew there were some smart aero owners out there. but dang! even though most of that was over my head... I got the gist of it. no matter the circumstance, the van will burn more fuel at 60 than 45.

what about horsepower? at 45 mph in drive, the van is at a different rpm than at 60 in drive. therefore the horsepower produced by the 4.0 should be more or less at the higher RPM. i am thinking more.

does the increased drag of 60 mph wind resistance defeat the increased horsepower produced at the higher rpm?

and actually I was surprised to know that the wind resistance at 60 was only 300 pounds. i was expecting much more! like over 1000 pounds.

do I subtract that extra 300 pounds from the tow rating of the van?

 

Manufacturers base tow capacities on weight, not wind resistance, however your suggestion is very interesting. How does the lifespan of a transmission change if one tows at 60 mph versus 70 mph (forgetting the effect of increased RPM for a minute and focusing on the effect of overcoming the added wind resistance? Tire capacity, for example, is speed based and has to be derated as speeds increase. Why not other components?

I think the preceding comments regarding resistance when towing are excellent. Few folks realize the relationship between speed and wind resistance is exponential.

Steve

 

Drag force is a square function of speed; that's a matter of physics. Therefore, drag power is a cubic function of speed. That means, given all else being equal, if you drive a given distance at 120 mph vs 60 mph, you expend 8 times the energy going the faster speed, just from the wind resistance. Other forms of friction vary by speed as well, so their contributions will further consume more energy.

As for parts longevity, in general, wear rate goes up with the square of the speed, assuming the load isn't also increased. When you drive faster, the increased drag automatically increases the load on all the parts that are pushing the car (but it won't be a whole order of magnitude). So yes, if you drive a little faster, you will wear out parts faster. By the way, you should not be going more than 60 mph while towing.

Here is a simple way to roughly figure out drag force while driving. If your car gets about 20 mpg at 60 mph, it is taking about 30 hp to push through the air at that speed. 30 hp is 30 * 750 ft-lbf/second or about 22500 ft-lbf/s. Divide by 60 mph, or 88 ft/s, is about 257 pounds drag force. (Tung was very close in his estimate.)

 

That's really interesting and I appreciate your time in posting this information.

Steve

 

The wind resistance is very valid to fuel consumption & thats why manufacturers try to design cars with the lowest "drag" coefficient. With towing, weight is also working against you, but cant help too much about that except to keep the weight down as much as possible.
However the engine is in effect a huge fluid pumping machine but the power output is not always linear with engine RPM.s. There are areas in the engines dyno curve where gas & air are being pulled in but the power output is not necasarily proportional to the fuel /air volume going in! Bearing that statement in mind there are indeed areas in the engines power curve which are more economical to travel in.
Cant say for sure where this range is on the 4.L engine but believe it is there somewhere!!

 

I am not trying to give a specific value, but the most widely published RPM figure in the consumer press has been between the values for peak torque and horsepower. Perhaps you know why that is; I do not.

If you have a reference covering this subject, I would appreciate learning of it.

Steve

 

Unfortunately, where the engine operates at its greatest efficiency is not always the speed at which you want to drive. It's close on the 4 liter v6; the torque peak is around 2400 rpm, while the power peak is around 4000 rpm. If the peak efficiency is somewhere in between there, that would put the vehicle speed way up around 70 or 80 mph, where the aerodynamic drag really works against it. So most of the time, you drive at a speed that's below the engine's peak efficiency. This also introduces throttling, another form of inefficiency.

This is where diesel engines have big advantages; they do not have throttles, and they are usually operated right at their peak efficiency range.

 

This has been most informative thread. Most of all I think it sheds light on my wife's South Korean import's ability to consistently achieve a higher mpg at 70-75 than at 60-65mph.Not much; almost a couple of mpg on I-95 to Florida and back to DC. Less in rolling terrain of I-81 to Tenn and back

I guess its sweet spot in all that math you guys were working with would be about 70.

 

I work on diesel engines for many years. Newer electronically controlled diesel engines with variable turbine geometry (VGT) have wider "green tacho area" and when I drive semi at 40 mph and at 60 mph I have no notisable MPG gate. But 30-40000 kg semi have 400-600 HP engine, and 32-38 l/100 km. so smaller truck of 2000-3000 kg weight will be hight MPG with 20-30 HP engine..... who will drive so slow pick-up truck?

 

The previous two statements which you are responding to...

1. "Assuming the drag coefficient is about 0.5 for that ugly trailer, that comes out to about 220 Newtons/square meter or about 4.6 pounds per square foot. The wind drag on your entire vehicle comes to about 300 pounds, give or take 100%"

2. "But of course, we're neglecting road friction. When you add that in, the actual number could be twice as much."

...are largely incorrect. The first statement is incorrect because it assumes the air velocity that the trailer sees is the same velocity the vehicle towing it sees. Generally speaking this is not true. The towing vehicle bears the brunt of moving the air out of the way of the trailer so that it creates a partial vacuum or draft behind it (similar to what road bicyclists and NASCAR drivers exploit). This means that the trailer will only create a fraction of the drag it would otherwise create if it were being pushed (instead of pulled) down the road by the tow vehicle (and was exposed to the full brunt of the oncoming air speed). The only partial exception to this is if the trailer's frontal area is larger than that of the towing vehicle. If that were the case, those portions of the trailer (those being outside the wake of the towing vehicle) would produce drag roughly as described in 1 above. But still, even accounting for those portions that stick outside of the profile of the towing vehicle, the net drag created by the trailer is almost always less than that created by the tow vehicle for the simple reason that the tow vehicle shields the trailer from the full force of the oncoming wind.

As to statement number 2 above, if the trailer tires (and bearings) are in good condition, not overloaded, properly inflated and rolling on a normal asphalt or concrete road surface, the rolling friction at highway speeds will be much less than that created by aerodynamic drag. As velocity diminishes, the rolling drag will eventually become dominate, but generally only at very low speeds and/or only on very rough road surfaces.

 

Thanks, I understand what you are saying. The trailer essentially "drafts" the tow vehicle. It am just so used to folks overlooking the relationship between speed and aerodynamic drag. In the world of RVs, nothing is ever tested in a wind tunnel so in my industry we end up talking in the abstract because no one really has a clue what the real world values are.

I appreciate your comments.

Steve

 

I need to correct something I said earlier:

"Drag force is a square function of speed; that's a matter of physics. Therefore, drag power is a cubic function of speed. That means, given all else being equal, if you drive a given distance at 120 mph vs 60 mph, you expend 8 times the energy going the faster speed, just from the wind resistance."

If you drive at twice the speed from point A to B, you get there in half the time, so you would only consume 4 times, not 8 times, as much fuel doing it.

On a similar note, I have seen drag coefficient figures for the Aerostar stated as anywhere from a very decent 0.37 to a brick-like 0.5. Does anyone know which it is?