Talk:Thrust-to-weight ratio

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On the Moon

"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)[reply]

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)[reply]

--

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)

--

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)[reply]

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)[reply]

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)[reply]

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)[reply]

JPG tabular data

Some user seems to be trying to add a table of data that they made in a spreadsheet to the article in jpg form. This form of data would be very difficult to change/expand/correct, and is completely out of keeping with the wikipedia's style.

Given that, I'm planning to remove it each time it appears.- (User) Wolfkeeper (Talk) 16:29, 11 March 2009 (UTC)[reply]

The table of data appears to be based on information sourced from www.fighterplanes.tk. This website contains many pages and it is not sufficient to cite the home page and hope this will constitute adequate verifiability of the data. Dolphin51 (talk) 22:24, 11 March 2009 (UTC)[reply]

Calculation : True Ratio

Anyone care to clarify the "It is a true ratio" line in the Calculation heading means? I can find no comparable use through searching around. As such, barring a link / further explanation in the article, I recommend its removal. Groxx (talk) 22:24, 29 May 2010 (UTC)[reply]

I changed it to "It is a dimensionless quantity.", it seems that was meant.--Patrick (talk) 00:23, 30 May 2010 (UTC)[reply]

Weight

This article refers to weight as a force in the definition but as a mass in the tables. Which is it?--208.54.14.122 (talk) 04:34, 3 July 2010 (UTC)[reply]

Thrust and weight are both forces. The weight of an aircraft or engine whose mass is known is easily calculated by multiplying the mass by the acceleration due to gravity. See the example given under the heading “ Examples”. Dolphin (t) 07:59, 24 December 2019 (UTC)[reply]

Low power examples

I think it would be interesting to have examples of very low ratio aircraft such as ultralights and early models such as the Wright Flyer. How low can the ratio be and still have a useful aircraft? Binksternet (talk) 18:46, 5 December 2010 (UTC)[reply]

That would be a glider. Zero!- Sheer Incompetence (talk) Now with added dubiosity! 20:02, 5 December 2010 (UTC)[reply]

For taking off at sea level and climbing without thermals... Binksternet (talk) 20:09, 5 December 2010 (UTC)[reply]

Powered glider. The L/D ratio is about 70, and they add minimal engine, so T/W would be about 1/70.- Sheer Incompetence (talk) Now with added dubiosity! 21:13, 5 December 2010 (UTC)[reply]

Seventy is excessive. Wikipedia states that the Fournier RF-5 and Scheibe Falke motor gliders both have L/D ratio of 22:1. The Schempp-Hirth Nimbus-2 has a much higher aspect ratio of 28.6 and its L/D ratio is 47.5. Perhaps the best-performing powered sailplane in commercial production at present is the Schempp-Hirth Nimbus-4M with a wingspan of 26.5m and aspect ratio of 38.8. Its L/D ratio is stated to be 60:1. The Nimbus-4M has a Rotax engine of 33kW power, so not a lot of excess thrust. Dolphin (t) 21:59, 5 December 2010 (UTC)[reply]

Thrust/Weight Ratio table: Calculation Error

if I may ask something being a mathematical dummy: if I do the maths for the following engine

- Pratt & Whitney F119[22] 1,800 3,900 91 20,500 7.95 (data see article) -

using the given formula t/w = (91 kN / (1800 kg * 9,807 kg/m²))*1000 my result is 5.16 and not 7.95 ... where's the error please? thanx! --HilmarHansWerner (talk) 19:54, 12 February 2013 (UTC)[reply]

You are correct. There is a problem with the data for the J58 and the F119. I have highlighted this on the Template Engine Thrust To Weight ratio Template_talk:Engine_thrust_to_weight_table

146.90.210.170 (talk) 17:54, 15 March 2014 (UTC)[reply]

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What is the significance of thrust to weight in a bare engine?

Thrust to weight for aircraft are shown for the entire vehicle, including fuel. For example take-off at sea level at max weight. This seems a very useful measurement. But for spacecraft, the thrust-to-weight figures are for the bare engine. Since the fuel type provides a specific impulse, what is the point of comparing sea level thrust if you don't include the rest of the launch vehicle and its fuel required to get to a particular trajectory? I can't get my head around how thrust-to-weight provides any insight for a spacecraft, once you know specific impulse. DouglasHeld (talk) 19:10, 2 September 2018 (UTC)[reply]

A picture is worth a thousand words

for example this one 85.193.247.94 (talk) 23:41, 23 December 2019 (UTC)[reply]

F-35

The article of the F-35 states "Thrust/weight: 0.87 at gross weight (1.07 at loaded weight with 50% internal fuel)", that's different than what the table in this article here says. Something is not right in both. Shouldn't the TWR be higher if the plane is empty and go down to more load? Why 0.87 at gross weight of 16t and more at "loaded weight", whatever that is, probably not "max take-off weight". The figures in the table here don't fit to this, TWR seems way too low. Also where is the 177kN figure from? The F-35 articles speaks of a 125kN engine, 191kN with afterburner. --OBrian (talk) 12:26, 15 July 2022 (UTC)[reply]

Recent addition regarding calculation of thrust-to-weight ratio

Zaereth recently made two good-faith edits to insert new material about aircraft weights, and calculating thrust-to-weight ratio. I consider this new material, at least in its original form, is unsatisfactory and so not suitable for Wikipedia.

The cited source for all this material is a NASA Technical Memorandum about trends in fighter aircraft. Combat aircraft in general, and fighter aircraft in particular, represent a very minor element of aircraft design and operation. Wikipedia is not a military encyclopaedia so its content and the sources on which it is based, should reflect the broadest sweep of aeronautics as evident in civil aviation. I would have no objection if a small element of military engineering is used as a minor appendix to the body of material on the subject.

The new material includes ... and even changes during flight operations but, apart from “munition load”, no example is given to explain why thrust-to-weight ratio might change during flight. (The primary reason T/W ratio changes in flight is variation in maximum power/thrust due to changes in pressure altitude and temperature.)

An explanation is given for “Empty weight”. This explanation looks like Operating empty weight.

An explanation is given for “Combat weight”. This is inappropriate. Thrust-to-weight ratio has a much broader application than just combat aircraft.

An explanation is given for “Maximum takeoff weight” but it is incorrect. MTOW is not “The weight of the aircraft when fully loaded ...” It is a weight determined during the certification process and published as a limitation that must be observed by everyone associated with operation of a type and model of aircraft. See Maximum takeoff weight. Dolphin (t) 13:56, 14 June 2024 (UTC)[reply]