Talk:Thrust

Rocket boosters

Are we sure that we got the space shuttle thing right? Holly goosh. Thrust is for babies. Because it seems to only have an acceleration of 7.2 m/s^2. Thats kind of slow isn't it? Even though one minute later it will be going at 1500km/h, but that would take a minute! Surely space shuttles accelerate faster than that. Pseudoanonymous 23:11, 15 May 2006 (UTC)

Information on Thrust

I have been listening to an exchange about what thrust is. The technical definitions do not help, so is thrust the force of the gases pushed through a jet engine and out the exhaust, or is it the force of the gases against the front of the engine.

An explanation would be greatly appreciated as soon as possible. (Its getting pretty heated here)

Thanks in advance for anyone's response or everyone's!

Cnfusd in MO Fshngoddes 10:31, 25 May 2007 (UTC)

Inaccurate Diagram

The diagram showing lift, weight, thrust and drag is not entirely correct. In UNACCELERATED flight all OPPOSING forces are equal. The diagram shows thrust being greater than drag, thus the aircraft must be accelerating. The article should note this or replace the diagram with an accurate one. Also, a diagram that shows the poper relationship between the 4 forces should show lift and weight being greater than thrust and drag. 163.118.44.134 18:22, 12 June 2007 (UTC)

Wrong scope of the thrust notion

If you begin from disambiguation, you will screen off fictional personages, pop song groups etc. and pick up physics. But inside physics you get straight to fluid dynamics and aircraft. I understand that this is because your fluid dynamics team is in charge. However, thrust forces occur in ground-based machinery, etc., and your article does not cover this matter. I have a question on thrust forces in structural analysis of concrete reinforcement and I failed to get even a definition. Is it just any force perpendicular to a plane? No answer in wiki. —Preceding unsigned comment added by 83.237.5.39 (talk) 09:10, 15 March 2009 (UTC)

Someone please explain 'pounds (lbs) of thrust'

Thrust is regularly quoted in pounds . i.e. A rocket engine has 10,000 lbs of thrust. In layman's terms what does that mean? Does it mean that if the rocket weighed 10,000lbs and the engine was generating 10,000lbs of thrust the rocket (pointed vertically) would not rise off the launch pad? I would love if someone would explain in simple terms the relation between pounds and thrust, with some simple examples like the previous rocket example.--Trounce (talk) 15:52, 13 April 2009 (UTC)

As you know, thrust is a force. In the SI system of units, force is measured in Newtons. In the old Imperial system of units, force is measured in pounds, often abbreviated lbf. See Pound (force). (The pound was also a unit of mass, often abbreviated lbm. See pound (mass).) A force of 1 lbf is equal to approximately 4.448 N. See Weight.

If a rocket weighs 10,000 lbf and it generates a thrust of 10,000 lbf, it will be able to support its weight at the earth's surface but unable to launch itself vertically upward. However, on the surface of the moon it will easily be able to launch itself vertically upward.

$W=mg\;$

where W is weight

m is mass

g is the acceleration due to gravity

Of course, it is essential that W, m and g are expressed in a consistent system of units such as Newtons, kilograms, and meters per second squared.

Under conditions of standard gravity (g₀ = 9.8 m/s² = 32.2 ft/s²) an object of mass M lbm has a weight of M lbf. However, when conditions differ from standard gravity (such as at the poles, or up a mountain, or in orbit around the earth, or on the moon) an object of mass M lbm has a weight different to M lbf.

lbf and lbm are not consistent units with respect to the above equation. When using the Imperial system of units where W is measured in lbf, m is measured in lbm and acceleration a is expressed in feet per second squared, the following equation must be used.

$W=m{\frac {g}{g_{0}}}$

where g is local acceleration due to gravity (in feet/sec²)

g₀ is standard gravity (32.2 ft/sec²) Dolphin51 (talk) 23:49, 13 April 2009 (UTC)

Excellent, Dolphin51 you've sorted it for me! As soon as I read "If a rocket weighs 10,000 lbf" it clicked with me. I was confusing pound force with pound mass with regard to weight. I forgot that weight is a force. For "weight" I normally use kilos, which is a unit of mass and that led me to think pounds of thrust was pounds mass!

Am I right in thinking that in a perfect world we'd use Newtons instead of kilos for weight? (and that would have saved my confusion from that start)--Trounce (talk) 15:15, 22 April 2009 (UTC)

Yes, in a perfect world we would all talk about weight in newtons, and the kilogram would only be used where we were clearly talking about mass. However, old habits die hard and we often talk about weight in kilograms, leading to confusion, particularly among students.

Some authors attempt to make it a little more rational by using the abbreviation kgf to indicate a weight (ie a force). The unit kilopond is no longer in widespread use, but it is the weight of a mass of one kilogram under conditions of standard gravity, and therefore equal to one kgf. Dolphin51 (talk) 23:07, 22 April 2009 (UTC)

Thrust to Power

Hello, I've found the section Thrust to Power too short and succinct for the casual reader, so I wrote a new section explaining the differences between Thrust and Power in a more didactic way. Feel free to review and merge as you see appropriate, and if you guys don't like the rewritten section feel free to remove it. Thank you!--Fbastos7 (talk) 04:34, 10 May 2009 (UTC)

Confidimus Propulso (In Thrust We Trust) —Preceding unsigned comment added by 41.6.55.31 (talk) 16:33, 21 June 2009 (UTC)

Hi,

When a jet is hovering, it is using energy. It might not be useful energy, but besides that, if an object is using energy during a period of time, it is also using power. I know the difference between using and producing, but what i try to say is that thrust can not only be used for forward energy, but also for climbing. The jet is not using power when it is falling at 9.8 m/s/s so when it is hovering, it uses 9.8 times its mass of energy per second, so the power is 9.8 times its mass. —Preceding unsigned comment added by 131.155.66.166 (talk) 10:19, 8 July 2009 (UTC)

Hahaha, yeah, a hovering aircraft is a tough case :^) That's why I had the idea of tying the aircraft to a tree, as zero speed is a great area for discussion, and clearly distinguishes the difference between power and thrust. What you're calculating there, as 9.8 times the mass, is the force that the engine is making -- that is, the thrust. So, that formula measures the thrust of a hovering aircraft, not the power. I noticed that students sometimes say that Power = Force / Time, but I guarantee, that's incorrect.. Power = Energy / Time.

Now, complex mechanical system have a lot of energy transfers all over the place, plenty of moving parts and all sorts of forces being applied. You can measure power and energy on many different ways. But engine makers want a simple number that you can measure, publish and rate their engines, so engineers simplify. Engineers usually identify the "useful" energy being produced by the engine. What "useful" means is completely arbitrary -- it can be generation of electricity, heating up of an oven, generation of microwaves in a microwave oven, or moving something around as in motor engines.

So, back to the hovering aircraft. That is the same situation as a horse pulling a cart. We want to calculate the power produced by the horse on the cart; that is, just like the aircraft, P=Tv. That is not all the power and energy transfers that the horse is involved with, but is the useful energy transfer to the cart (or the useful power on the cart). If it happens that the cart has a broken axle and won't move, however much force the horse spends, then the power the horse produced on the cart is zero -- although the horse would disagree, because it probably got very tired trying to pull a broken cart.

So, say that you got pity on the horse and wanted to calculate all the energy the horse spent on everything else. Then you add all energy transfers: breathing, sweating, noise-making, hoofs sleeping on the floor, etc.. That gives all energy transfered by the horse into the cart + environment. That is certainly greater than zero, even if the cart doesn't move. But it is not useful work on the cart. Similar thing with the jet engine... if you want to calculate all the energy transfers between the engine and the environment, you'll get a much larger number than P=Tv (it will include the air blown around, sound, heat, vibration, etc...). But that is not useful work. These are just requirements (energy loss) needed for the engine to produce useful work. If the aircraft is hovering, it is not changing energy... no energy change, means no work on the aircraft... which means the engine is producing zero power on the aircraft (even if the engine is consuming a lot of power and wasting most of it to blow air around as a condition of keeping the aircraft stationary).

Cheers Fbastos7 (talk) 17:21, 7 October 2009 (UTC)

The "Thrust to Power" section of this article contains misleading and outright wrong information. The formula presented for Power is correct, however the following arguments are not. To begin, the thought experiment regarding aircraft tied to a tree is a poor choice, as it creates a physically incorrect mental image of flying tethered aircraft. Neither plane would be capable of flight without relative airspeed to create flow over their respective lifting surfaces. Ignoring this, we'll go point by point and break down what is wrong with the current article:

> if a jet aircraft is at full throttle but is tied to a very strong chain to a tree, then mathematically the jet engine will produce no power. It > produces a lot of thrust, but it doesn't move, so it performs no work and therefore has zero power.

Saying that the jet engine performs no work in this situation is incorrect. Assuming that the engine is operational, there will be a jet of incredibly hot air blasting out of its rear end. The high velocity and temperature of this air reflect the fact that the engine indeed has done work. It would be correct to say that the engine has done no work *to move the aircraft,* but that much should already be obvious since the aircraft isn't moving. The author is correct in noting that the engine is producing a thrust force.

> Conversely, a piston engine produces work by turning the propeller, that is, by creating torque on the propeller. It doesn't create any force on the > plane by itself; so a piston engine has no thrust. The propeller is the entity that accelerates the air and creates a force in the airframe -- so the > propeller is the entity that produces thrust. If you tie a piston aircraft tied to a very strong chain at full throttle, then its engine will have zero > thrust and full power.

The distinction that the author makes here between the engine and the propeller is meaningless. The engine produces thrust force by turning a propeller. The engine is still generating thrust force. If you tethered a parked piston engine aircraft to a tree and then started the engine, there would be a tensile force in the line equal to the thrust force produced by the engine. It's easier to think about the work done by a spinning shaft (Moment*angular displacement) so perhaps that is why the author is willing to believe that the engine is doing work in this case (which is true). Just as in the jet case, the engine does no work to move the aircraft.

I think that the author's confusion comes from a misunderstanding of work and power. It's true that the portion of a jet engine's power output which goes toward forward motion can be calculated using the P=Tv formula he/she presents. It is incorrect, however, to say that an engine which is not working to propel an aircraft is doing no work. Both engines produce thrust despite the aircraft being tethered to the tree.

The article as written seems to be a misinterpretation of this article: http://www.aerospaceweb.org/question/propulsion/q0195.shtml which oversimplifies the underlying issues, but correctly outlines why turbojets are typically discussed in terms of their thrust force and piston engines are discussed in terms of their shaft power output.

A better way of dealing with this topic in the Wikipedia article would be to eliminate the discussion of the planes tethered to the tree and replace it with a discussion which explains why it is convenient to rate piston engines and jet engines differently. Propeller efficiency (in trying to determine thrust of a piston engine) and difficulty in calculating work done on the air that passes through a turbojet (in trying to determine the static power output of a turbojet) should be identified as the oft unknown factors which make comparisons between the two types difficult to make.

I appreciate the remarks (albeit the dozen times you wrote I "mis-something"), but I didn't invent the formula P=Tv. In this formula, v is the speed of the airplane -- not the speed of the air being moved around. If v=0, then P=0.. there is no way to argue around that. If the jet engine is blowing a lot of air around, is making a lot of sound and heat, and is vibrating the aircraft... none of that matters when you are talking about the Power that a motor with a Thrust rating produces. Now, if your jet engine was a leaf blower, it would be different. A jet engine is a motor, its purpose is to move the airframe around... if the airframe doesn't move, then the work is zero and the power is zero. Similarly, an internal combustion engine's purpose is to move the crankshaft.. if the crankshaft breaks and doesn't move any more, the engine is producing zero power (even if the motor is at full throttle).

Now, if you don't believe in that, then please write your formula linking Power to Thrust for a jet engine.

Fbastos7 (talk) 16:03, 7 October 2009 (UTC)

Like I said above, which is in agreement with your assertion, a jet engine on a tree-tethered aircraft does no work *to move the aircraft* essentially by definition since the aircraft is stationary. The statement that a restrained jet engine can't do work *to move itself or anything it's attached to* is trivial. My point is that it is incorrect to say that this engine does no work without the qualifier above.

Power output can be related to thrust force for a turbojet independent of the motion of any aircraft to which the engine may be attached, as I demonstrate below.

First we assume that the engine operates ideally (the compressor and turbine function quasi-statically and adiabatically, fuel combustion is essentially the addition of heat at constant pressure, and exhaust air cools to ambient temperature at constant pressure). Also for simplicity we can assume that the mass flow rate of fuel into the combustion chamber is much smaller than the mass flow rate of air into the engine and that the exit pressure of the working fluid is equal to the ambient atmospheric pressure.

Note: I've written each term out in full to reduce ambiguity. The variable cp is the constant pressure specific heat capacity of air.

Thrust force generated by the engine under these assumptions will be:

1. F = [mass flow rate of air into (or out of) the engine] * (air exit velocity - air inlet velocity)

Work done per unit mass flow into/out of the engine will be:

2. w/(unit mass) = [heat rejected by the system as exit air cools to ambient temperature (negative since this heat exits the system)] + [heat received by the system when fuel is combusted (positive since this heat enters the system)]

Which can be broken down as follows:

3. (heat rejected by the system as exit air cools to ambient temperature) = (specific enthalpy of air at ambient conditions) - (specific enthalpy of air at the turbine exit) = cp*(ambient atmospheric temperature - turbine exit temperature)

and

4. (heat received by the system when fuel is combusted) = (specific enthalpy of air at the turbine inlet) - (specific enthalpy of air at the compressor exit) = cp*(turbine inlet temperature - compressor exit temperature)

Which gives:

5. w/(unit mass) = cp*[(turbine inlet temperature - compressor exit temperature) + (ambient atmospheric temperature - turbine exit temperature)]

Power output will then be equal to:

6. w/(unit mass) * mass flow rate of air into or out of the engine

Combining equations 6. and 1. allows you to write:

Power output = Thrust Force*cp/(turbine exit velocity - inlet velocity) * [(turbine inlet temperature - compressor exit temperature) + (ambient atmospheric temperature - turbine exit temperature)] —Preceding unsigned comment added by 98.216.106.112 (talk) 23:07, 12 October 2009 (UTC)

My friend, Wikipedia is an open effort. If you feel like it, feel free to write those equations in the page, but you'll get every reader confused. You are measuring the work that the engine is doing on the air, not on the airframe. That is a fine calculation, but it is not useful work. Engines are rated by their useful work, not by the work they do on their surroundings, or the other energy transfers that are not involved on their purpose. For example, your equation also works for a candle on a table top -- so one can conclude that "the candle is producing so much power to move the air". But, as an engine, the power rating of a candle is zero. I don't know what else I can argue: I agree with you that jet engines heat up and move air around, but I don't agree with you that jet engines are rated by how much they heat up and move the air around -- that would work well for a leaf blower, though.

Thanks. Fbastos7 (talk) 03:28, 1 November 2009 (UTC)