Hi rudderrudderrat:
Yes I agree with the quote you provide from the Navy document. Do you understand that the Navy document does not mean what you had previously said about THP? The following sentence is not the same:
"We agree that we are not creating any useful THP - but we are most certainly accelerating a mass of air and have given it a relative velocity - thus we are developing THP. Otherwise where is the energy of the fuel we are using going to?
The engine is producing BHP using the energy of the fuel. "THP" means "usable thrust", as the Navy document said - not "useful (usable) THP".
Does that make sense?
Re: oggers' helicopter hover situation - I haven't studied any aerodynamic stuff regarding helicopters, however, I believe that when the helicopter is in a fixed position over the ground (in a hover), the THP will be zero. The engine will definitely be creating lots of BHP(SHP) and burning lots of fuel to produce the thrust that is keeping the helicopter in the hover - but I believe THP will be zero. Work will be done to lift the helicopter off the ground into the hover position which will obviously take a certain amount of THP.
This is my reasoning. First of all, I'm assuming that the same concept of THP (ie: with regard to the airplane's performance) is applied to the helicopter scenario. As I've stated, THP is related to the
performance of the aircraft. First assume that the helicopter is a small cardboard box that is empty. It is resting on 4 poles situated at the four corners of the box so that you are free to place your hand underneath the box and lift it. When the box is resting on the 4 pole platform I think we can agree that there is no work being done on the box. Then you place your hand underneath it and apply a 10 Newton force in the direction of the normal (vertical) axis. Let's assume it takes 100 Newtons of force to overcome the force of gravity on the box. So increasing the force of your hand on the box all the way to 100 Newtons will essentially transfer the weight (mass x gravity) of the box to your hand. At the point where you are applying 100 Newtons of force, the box still hasn't moved. Work = force x distance. Therefore, no work has been done on the box. If no work has been done on the box, then power equals zero. Power = work / time.
Then you apply a 150 Newton force for 1 second before reducing the force to 100 Newtons. That accelerates the box upwards and then brings the box to rest again at a position that is a certain distance from the original position (resting on the 4 pole platform). Since you changed the position of the box you did work on the box. And since you did work on the box, power was used to move that box. But once it comes to rest again, it goes back to the same condition - zero work and zero power on the box.
It's the same with the helicopter. I'll disregard ground effect for this thought experiment. The pilot increases the thrust to get it off the ground and then reduces it so that he comes to a hover a certain distance off the ground. He still needs some force that will oppose the mass x gravity of the helicopter and the thrust provides that. The engine is doing lots of work to generate that thrust - but the thrust is not doing any work on the helicopter when it's in the hover! No work = no power!
-------------------------
oggers.. why don't you prove me wrong?! If you've taken all these exams and gotten educated on these topics, do you have any references that you could provide that proves me wrong? So far, you've been very adamant in saying I'm wrong but you've provided no counter argument to the references that I've provided... which leads me to my next point...
You say you don't disagree with any of the references I've provided. That's progress. But you say that they don't support my point. Can you provide evidence that my point is different than what is said in the references I provided?
Because – as I pointed out before - that Navy stuff is correct. But it doesn’t support your point, it is redundant in that context.
Well this is what you said:
"For the sake of clarity I agree that THP will not equal BHP because of prop efficiency (I should have been more careful to specify SHP in front of a pedant) and frictional losses in the intermediate gearing. But that is just semantics..."
But you actually don't understand what that 'prop efficiency' really is! Here is the equation for propeller (propulsive) efficiency:
http://i.imgur.com/NXLV3.png
And you say that's just semantics... haha
Then you said:
"But that does not mean a stationary aircraft can do no work and therefore produce no THP!!! The work is done by moving the mass of air one way and the mass of the Earth a tiny imperceptible amount the other way."
But the 'Navy stuff' clearly showed that you were in fact creating zero THP when the aircraft was stationary! So you obviously don't agree with the 'Navy stuff'. You're still not getting that we're talking about aircraft performance. I'm not sure how many times I should repeat "aircraft performance" before it sinks in. Of course you're doing 'work' on the air because you're moving it backwards. And I'd like to add emphasis to the part where I said:
"doing 'work' on the air". You are NOT doing any work on the airplane!
What's interesting though is that you said in your most recent post:
"Are you telling me that when I was releasing bolts you were in the engine? BTW everyone knows that 'no work is being done on the vehicle'. But: work...is...being...done...on...the...air.
So you do understand that no work is being done on the 'vehicle' and yet when I say that no work is being done on the airplane when it's stopped (therefore there is no power), you have told me that I'm wrong. Hmm... haha
I've never disagreed that the engine is doing work (BHP) but I have been very clear in describing that if the 'vehicle' is not moving, then the THP is zero.