4 forces acting on unaccelerated, straight&level flight?
Thread Starter
Join Date: Jan 2008
Location: Kabul, Afghanistan
Age: 40
Posts: 274
Likes: 0
Received 0 Likes
on
0 Posts
4 forces acting on unaccelerated, straight&level flight?
As FAA handbook of aeronautical knowledge states that in a straight and level unaccelerated flight the Thrust is equal to Drag and Lift is equal to Weight, but the lift and weight is greater than thrust and drag. I have been asked why lift & weight is greater than Thrust and drag?
Join Date: Feb 2006
Location: Mel-burn
Posts: 4,875
Likes: 0
Received 0 Likes
on
0 Posts
I'm happy to be corrected.....
Think of it like this-
If your plane was a helicopter and the prop was at the top, it would NOT have enough lift/thrust to take off. Therefore lift / weight MUST be greater than drag and thrust.
As for WHY they are greater... gravity.
Think of it like this-
If your plane was a helicopter and the prop was at the top, it would NOT have enough lift/thrust to take off. Therefore lift / weight MUST be greater than drag and thrust.
As for WHY they are greater... gravity.
Join Date: Feb 2007
Location: Home
Posts: 100
Likes: 0
Received 0 Likes
on
0 Posts
If the earth was in a vacuum, the prop wouldn't work!! The engine wouldn't neither..... sorry couldn't help myself.
Well anyway the aircraft still has mass and inertia to overcome.
Well anyway the aircraft still has mass and inertia to overcome.
Sounds a bit simplistic perhaps? Lift is going to equal weight no matter how fast you're going S & L, but if an aircraft with a great big donk sets a thrust greater than the equivalent of its weight, it'd just keep accelerating - drag builds up with V squared until it equals thrust, and both of them are bigger than L or W. Wouldn't work for your average bug smasher but.
Join Date: Aug 2007
Location: Home
Posts: 46
Likes: 0
Received 0 Likes
on
0 Posts
An airfoil requires a certain amount of velocity of airflow to create the pressure differential between the top and bottom surface required to equalise the lift and weight components. The actual difference in pressure is quite small. Aircraft are designed aerodynamically therefore less drag equals less thrust required.
Anyways, it is much easier to move horizonally than vertically.
So, you may notice that your rated output of thrust is quite a bit less than your weight (if you are able to convert the two into a single unit of measurement).
E.G. A mate and I only yesterday were considering thrust outputs on a 747-300 with RR RB211-524D4 engines which had a rated thrust of 51980lbs minimum times 4 engines.
Now if anyone figures something different I'm happy to listen and be corrected if I'm wrong.
This equates to a little over 94 tonnes of thrust total. Awesome.
The 747 MTOW is almost 400 tonnes. Awesome.
Hope that helps.
Anyways, it is much easier to move horizonally than vertically.
So, you may notice that your rated output of thrust is quite a bit less than your weight (if you are able to convert the two into a single unit of measurement).
E.G. A mate and I only yesterday were considering thrust outputs on a 747-300 with RR RB211-524D4 engines which had a rated thrust of 51980lbs minimum times 4 engines.
Now if anyone figures something different I'm happy to listen and be corrected if I'm wrong.
This equates to a little over 94 tonnes of thrust total. Awesome.
The 747 MTOW is almost 400 tonnes. Awesome.
Hope that helps.
Left Drag ratio?
The simple answer is your average light airplane weighs around 3000 lbs, therefore needs 3000lbs of lift.
However, it DOES NOT have 3000lbs of drag, therefore does not need 3000lbs of thrust.
Enter the L/D ratio, which for a half decent airplane is aroundabout 10/1, with the usual variables of manufacturer, flap deployment and all that other rubbish.
So, if your AFM says your L/D ratio is 10/1, and you weigh 3000lbs, your thrust and drag components are 300 lbs each, in S&L flight, and that is why the four forces are not equal to each other.
However, it DOES NOT have 3000lbs of drag, therefore does not need 3000lbs of thrust.
Enter the L/D ratio, which for a half decent airplane is aroundabout 10/1, with the usual variables of manufacturer, flap deployment and all that other rubbish.
So, if your AFM says your L/D ratio is 10/1, and you weigh 3000lbs, your thrust and drag components are 300 lbs each, in S&L flight, and that is why the four forces are not equal to each other.
Lift/Drag ratio of the wing...
You build an aeroplane that weighs X designed to carry Y people etc. Physics shows that Lift has to equal Weight. The ratio of Lift to Drag is defined by the characteristics of the wing...say 10:1...also gives approx glide ratio. Thrust only ever needs to be "a bit more than" drag (over simplification) so those two forces will only ever "need" to be a small proportion of the Lift and Weight.
By the way, there are FIVE forces...CASA.
You build an aeroplane that weighs X designed to carry Y people etc. Physics shows that Lift has to equal Weight. The ratio of Lift to Drag is defined by the characteristics of the wing...say 10:1...also gives approx glide ratio. Thrust only ever needs to be "a bit more than" drag (over simplification) so those two forces will only ever "need" to be a small proportion of the Lift and Weight.
By the way, there are FIVE forces...CASA.
FWIW in S&L flight lift is slightly more than weight.
The reason is that it needs to counteract the downforce created by the tail, which is used to generate dynamic stability.
The reason is that it needs to counteract the downforce created by the tail, which is used to generate dynamic stability.
18-Wheeler - true, but the downforce from the tail acts downwards, increasing the weight of the aircraft in flight, (but making no change whatsoever to the mass of the aircraft).
Join Date: May 2007
Location: Aus, or USA, or UK or EU, or possibly somehwere in Asia.
Posts: 320
Likes: 0
Received 0 Likes
on
0 Posts
Lasiorhinus is correct.
even on small aircraft the stall speed increases with a fwd cog, that is why FAR 23 requires the stall speed to established (amongst other things) at the forward limit at MTOW. other handling issues are required to be investigated at the forward regardless cog.
Net lift will equal weight, ie subtract the downward 'stabilising' force from the upward lift produced by the mainplane(s) and fuselage and the net will be the same as weight, the airacrft just thinks that it is heavier though mass remains unchanged. Also consider that the Vh is reduced at fwd cog due to the increased drag due to the added trim drag of the tailplane and that the increased total lift (up and down) produces added induced drag, ergo slower speed for same thrust. aft cog will reduce Vs0 and increase Vh in all aircraft, the larger the aircraft the more quantifiable this becomes.
HD
even on small aircraft the stall speed increases with a fwd cog, that is why FAR 23 requires the stall speed to established (amongst other things) at the forward limit at MTOW. other handling issues are required to be investigated at the forward regardless cog.
Net lift will equal weight, ie subtract the downward 'stabilising' force from the upward lift produced by the mainplane(s) and fuselage and the net will be the same as weight, the airacrft just thinks that it is heavier though mass remains unchanged. Also consider that the Vh is reduced at fwd cog due to the increased drag due to the added trim drag of the tailplane and that the increased total lift (up and down) produces added induced drag, ergo slower speed for same thrust. aft cog will reduce Vs0 and increase Vh in all aircraft, the larger the aircraft the more quantifiable this becomes.
HD
18-Wheeler - true, but the downforce from the tail acts downwards, increasing the weight of the aircraft in flight, (but making no change whatsoever to the mass of the aircraft).
Join Date: Apr 2008
Location: Australia
Posts: 669
Likes: 0
Received 0 Likes
on
0 Posts
I think everybody is missing the point of the question.
There is no rule that says L&W must be greater than T&D. The two sets of forces are completely unrelated.
The T&D could be greater than the L&W of course, but that would be a rare aircraft to possess that amount of thrust - but note that there is no law of physics that says the T&D cannot be greater than the L&W.
Why are the main wheels bigger than the nosewheel? Same sort of question. They don't HAVE to be, but just usually are.
There is no rule that says L&W must be greater than T&D. The two sets of forces are completely unrelated.
The T&D could be greater than the L&W of course, but that would be a rare aircraft to possess that amount of thrust - but note that there is no law of physics that says the T&D cannot be greater than the L&W.
Why are the main wheels bigger than the nosewheel? Same sort of question. They don't HAVE to be, but just usually are.
Join Date: Apr 2008
Location: Australia
Posts: 669
Likes: 0
Received 0 Likes
on
0 Posts
You may want to have a little think about that statement.
Yes, completely separate and unrelated.
To prove that statement, consider an aircraft in unaccelerated, "straight and level" flight. Then, go about removing each of the forces, and see whether their removal has any effect on the other forces.
Of course, removing one force causes a change to the flightpath, which then causes the other forces to change, but note that it was the flightpath change that caused the changes to the other forces - NOT the loss of the original force. This is the proof that the forces are unrelated to each other.
To the original question that this thread is based on: it is possible to build an aircraft, where, in unaccelerated "straight and level" flight, the T&D are greater than the L&D. In fact I wonder if some military aircraft can already do this.
To build your own, obtain an engine with plenty of thrust, say one of the current generation of turbofan engines that can do 100,000 lbs of thrust. Then, attach it to an airframe that weighs less. Think of a tiny airframe, with a tiny fuel tank and that great big engine.
Take it for a flight and have the thrust set at maximum. You will be going fast, and if your airframe (and fuel) weighs less than 100,000 lbs (about 45,000 Kg) then your T&D is greater than your L&W.
Again I say, there is no law that says the L&W must be greater than the T&D.
I think this question has arisen from the fact that the former set is USUALLY greater than the latter.
Join Date: Dec 1999
Location: Melbourne VIC AUS
Posts: 116
Likes: 0
Received 0 Likes
on
0 Posts
FGD135
Yup, and at about M3 you would finally reach equilibrium - you would need supersonic drag to counter that thrust.
I think that what the FAA (or indeed our original poster) forgot to say was that the comment is applicable to GA light aircraft as commonly certified.
Yup, and at about M3 you would finally reach equilibrium - you would need supersonic drag to counter that thrust.
I think that what the FAA (or indeed our original poster) forgot to say was that the comment is applicable to GA light aircraft as commonly certified.