Talk:Thrust-to-weight ratio

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"for launch from the Earth's surface, for launch from the Moon it only needs to be more than 0.1654"

Thrust to WEIGHT ratio always need to be larger than unity. Moon's surface gravity 1/6 of Earth, but you still need 1N thrust to counter 1N of gravitation attraction. I think the editor is mixing up mass and weight. —Preceding unsigned comment added by 59.149.87.75 (talk) 10:47, 1 November 2007 (UTC)

[edit] aircraft performance entries

Wiki aircraft performance entries show the following in the thrust/weight: data 'F100 0.898; F110 1.095', but this article in no way explains what this meant!--Mrg3105 10:54, 28 July 2007 (UTC)

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I am not shure if I got it right, but this is what I know. The article tells the following:

> For a takeoff using pure thrust and no wings, the thrust-weight ratio for the vehicle has to be more than one [...]

This primarily means rockets/space craft. Other aircraft heavier than air use rotors or wings to gain height (airliners, helicopters etc.), plus some sort of engine and steering, to move horizontally. Airplanes don't need a thrust/weight ratio above 1 (see your F100 ratio). But a good ratio - and a high wing loading - is most important to modern fighter airplanes. VTOL airplanes like the Harrier do definitly need a thrust/weight ratio above 1 to start vertically. A ratio of more than 1 provides possibility to fly vertically for those aircraft, which means without the help of wings and their aerodynamic lift.

Also check out the F/A-18 Hornet - especially the "design" section in this article:

> [...] superbly maneuverable, owing to its good thrust to weight ratio [...], and check out this pic: a Hornet climbing - sthing a F-100 can't do I think. ;-)

Greetings, Andi, 13:31, 27 September 2007 (UTC)

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[edit] local gravitational field strength

The first and last paragraphs of the first section are contradictory: the first says the ratio is pegged to Earth's surface gravity; the last says it is not. —Fleminra 06:40, 1 October 2007 (UTC)

Actually, it's the other way around. Still a contradiction. It results from the fictionalizing of the concept to pretend that it really is a dimensionless ratio, when it is not. There is no acceleration of gravity involved in these calculations, but rather you need either the pretense that pounds-force will cancel out pounds (or equivalently, kilograms-force will cancel out kilograms), or you need to add a gratuitous fudge factor to the calculations, something not called for by the physics involved: the standard acceleration of gravity (a concept of metrology, not of physics). Gene Nygaard (talk) 18:12, 21 March 2008 (UTC)

For aircraft it doesn't matter much, since gravity doesn't vary. For rockets the standard seems to be to peg the local gravity to g0. Sutton (7th edition pg 442) says that: thrust-to-weight ratio F/Wg is a dimensionless parameter that is identical to the acceleration of the rocket propulsion system (expressid in muliples of g0) if it could fly by itself in a gravity free vacuum (and they define Wg to be loaded weight at sea-level.)- (User) WolfKeeper (Talk) 18:30, 21 March 2008 (UTC)

Actually, the last paragraph says "actual performance will often be affected by … local gravitational field strength," not that T/W is affected by local gravitational field strength, so never mind. —Fleminra (talk) 21:29, 21 March 2008 (UTC)