Aviation brain teasers

[TeeVee]

plane going west to east will be in the jet stream.

tried that. it's a hypothetical and there is NO wind

[TeeVee]

and the answer is..........the one flying west.

Eötvös effect.

any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, thanks to the Earth's rotation.

[guamainiac]

TeeVee, back in the 60's I got a 727 out of Honolulu straight into JFK. All the west bound flights had to stop for fuel in CA before going on; I asked and was told by a crew member that if could be done in a 727 using the jet stream.

I did catch that "no wind" thing later and was going to delete but figured it wasn't fair taking your "dirty laundry" off the line.

And where the heck did you dig that up? Esoteric as all get out.

[elaw]

and the answer is..........the one flying west.

Eötvös effect.

any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, thanks to the Earth's rotation.

Good call!

But one teeny thing: the original question is which airplane is using less fuel.

So the answer would be the *eastbound* one... as the Eötvös effect causes a reduction in the perceived force of gravity when traveling toward the east. Thus the eastbound plane will have less induced drag, thus requiring less power to remain airborne, thus consuming less fuel.

[Gabriel]

Eötvös effect.

any object co-rotating with the Earth at the equator has its measured weight reduced by 0.34 percent, thanks to the Earth's rotation.

Yes!!! And thanks for the name, I didn't know that it was called Eötvös effect.

and the answer is..........the one flying west.

Hmmmmm.....

Good call!

But one teeny thing: the original question is which airplane is using less fuel.

So the answer would be the *eastbound* one... as the Eötvös effect causes a reduction in the perceived force of gravity when traveling toward the east. Thus the eastbound plane will have less induced drag, thus requiring less power to remain airborne, thus consuming less fuel.

Correct, you are the winner and the prize goes to you, together with the obligation to present a new puzzle.

A bit of explanation of the Eötvös effect (non-mandatory reading)

In simple terms, the centrifugal force of something rotating around the Earth cancels a bit (or a lot) of the apparent weight, and the centrifugal force is greater the faster an object is rotating around the Earth, what an object is already doing if it's static because the Earth itself is already rotating around itself (at about 1000 MPH on the equator, which is about twice the cruise speed of a 777). So if you move "with" the Earth your speed adds to that of the Earth, and if you are flying in the opposite direction you have to subtract your speed to that of the Earth.

So the airplane flying "with" the Earth will have a lower apparent weight, will need less lift to bear that lower apparent weight, and will then have less induced drag (which is the drag of doing lift). We all know that a "lighter" airplane burns less fuel, right?

A bit more of detail:

If you moved fast enough around the Earth you would be in orbit, and apparently weightless. If you go high enough over the Equator, "fast enough" becomes "at the same angular speed than the Earth", and then you become a geostationary satellite. "High enough" provides for a higher tangential speed and a lower gravitational force.

But at normal airplane altitudes, the altitude itself is almost negligible because it is so small compared to the radius of the Earth.

So, a 777 at say 500 MPH flying "with" the Earth on the equator will have a tangential speed of 1000+500=1500 MPH, while one flying due West will have a tangential speed of 1000-500=500 MPH.

Because this centrifugal force goes with the square of the tangential speed, if 1000 MPH cancels 0.34% of the wight as TeeVee correctly said, 1500 will cancel 0.765% of the object's weight and 500 MPH will cancel 0,085% of its weight. That's a 0.68% less apparent wight of the plane flying "with" the Earth than the one flying "against" it, and a 0.68% less of induced drag, which at minimum drag speed is about half of the total drag, so 0.34% less thrust and, if the efficiency of the engine was linear with thrust, 0.34% lower fuel burn. (airplanes fly faster than minimum drag speed so the effect will be a bit lower than that)

[elaw]

Well I think 90% of the credit should go to Tee Vee - all I did was read the Wikipedia article!

As for another puzzle... give me a little time to see if I can come up with something good.

In the meantime a pretty lame one but let's see if anyone can answer it *without* using an online search engine (honor system): what is the product name (as opposed to model number) that P&W used for many of their early turbine engines such as the JT3 and JT8?

[Gabriel]

Well I think 90% of the credit should go to Tee Vee - all I did was read the Wikipedia article!

Actually, I think he did this "mistake" intentionally to avoid having to post a puzzle
(but I have no evidence)

[elaw]

Damn! I wish I'd have thought of that...

[Highkeas]

I used to work for BAC and one of the surprises once the Concorde became operational is that fuel usage was reduced slightly due to centrifugal force.

[TeeVee]

Actually, I think he did this "mistake" intentionally to avoid having to post a puzzle
(but I have no evidence)

who me????? i would never