Does the Thrust need to be greater than the weight?
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Does the Thrust need to be greater than the weight?
I have to say I only know the basics of how a plane works. But one thing I'm still wondering about; Does the Thrust generated by a plane's engines need to be greater than the weight of the plane, for it to move, and thus take off?
Looking at the specs for a Boeing 747-400, I see that it weighs 800,000, but the engines produce 63,300 lb of thrust (is this for one engines, or all 4?).
This may be a "Duh!" moment for you guys, but bear with me
Looking at the specs for a Boeing 747-400, I see that it weighs 800,000, but the engines produce 63,300 lb of thrust (is this for one engines, or all 4?).
This may be a "Duh!" moment for you guys, but bear with me
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The only time you need the thrust to exceed the mass (better term than weight) is if you want to go vertically upwards, i.e. like a space rocket or a military jet at an air display.
In steady flight, mass is balanced by LIFT not thrust. The opposing force here for thrust is DRAG
Hope this helps a little.
Try searching this site for 'Principles of Flight' or Google for oddles of good stuff.
Cheers,
The Odd One
In steady flight, mass is balanced by LIFT not thrust. The opposing force here for thrust is DRAG
Hope this helps a little.
Try searching this site for 'Principles of Flight' or Google for oddles of good stuff.
Cheers,
The Odd One
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Milt,
Nice one, but it'll only be form & parasitic, shouldn't be any induced drag as no lift is being generated (if the Zero Lift Angle is being 'flown')
Cheers,
The Odd One
Nice one, but it'll only be form & parasitic, shouldn't be any induced drag as no lift is being generated (if the Zero Lift Angle is being 'flown')
Cheers,
The Odd One
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.... hmmm, without giving it too much thought, I think the wing is rather indifferent to the path of the a/c relative to the earth, so will still be generating lift, whether or nor the a/c is straight and level, inverted, vertical, etc. (assuming a positive A of A) So say an a/c in a vertical trajectory would all things being equal, have the nose slightly off vertical so the 'weight' could counteract the lift vector, or be flying with a A of A of exactly zero. Where the hell is my copy of Kermode when I need it.
Anyone actually flown vertical have a better answer?
Anyone actually flown vertical have a better answer?
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I don't see why the wing would be producing any lift at all. Use the horizontal stab/elevator to reduce aoa to zero at the 90 degree point.
For Milt's response, I think he means TAS, as IAS would mean TAS increases in the climb.
Hawk
For Milt's response, I think he means TAS, as IAS would mean TAS increases in the climb.
Hawk
You fly a "Zero Lift" A of A which would be zero on an aircraft with a symetrical wing, or slightly less than zero on (i.e. a slight negative A of a) on an aircraft with a cambered wing.
As most aircraft have a positive angle of incidence, the deck angle would be slightly nose down, rather than pure vertical.
Yes it does, because drag decreases with altitude due to lower air density. Thus for a constant thrust, a constant IAS but increasing TAS would result. (However, as most types of engine reduce in power with altitude, this doesn't seem likeley!!)
As most aircraft have a positive angle of incidence, the deck angle would be slightly nose down, rather than pure vertical.
as IAS would mean TAS increases in the climb.
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Wizofoz, the IAS cannot cannot remain the same. Here is Milt's quote
“For the high performance aircraft climbing vertically at constant IAS the thrust has to be equal to the weight PLUS the drag!!”
If the thrust is equal to weight plus drag, then there is no excess thrust to accelerate the aircraft. Because to keep the same ias, then the TAS would have to increase. It is, of course, climbing up, into less dense air.
My post was to suggest that instead of “IAS”, that the use of “TAS” would be correct for Milt's post.
In Milt’s case, I believe the thrust would have to be greater than the weight plus the drag, in order to accelerate and keep the same IAS. Again, his applies to Milt's scenario.
Hawk
“For the high performance aircraft climbing vertically at constant IAS the thrust has to be equal to the weight PLUS the drag!!”
If the thrust is equal to weight plus drag, then there is no excess thrust to accelerate the aircraft. Because to keep the same ias, then the TAS would have to increase. It is, of course, climbing up, into less dense air.
My post was to suggest that instead of “IAS”, that the use of “TAS” would be correct for Milt's post.
In Milt’s case, I believe the thrust would have to be greater than the weight plus the drag, in order to accelerate and keep the same IAS. Again, his applies to Milt's scenario.
Hawk
thrust has to be equal to the weight PLUS the drag!!”
It is exactley the same theory as a normal, constant IAS climb in an aircraft with a flat rated engine. For the same thrust and IAS, TAS will increase with altitude.
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Ohhh - give me these engines to fly vertically
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Oz, reference your quote:
"Drag then decreases as the air density reduces, thrust is now greater than drag so the aircraft accelerates.
It is exactley the same theory as a normal, constant IAS climb in an aircraft with a flat rated engine. For the same thrust and IAS, TAS will increase with altitude"
If it is a constant ias climb (can we talk cas?) as you say, then how can drag decrease (compressibilty aside)?
Secondly, I know engines can be flat rated with respect to temperature. Can you explain this flat rating with altitude and speed? New to me.
Hawk
"Drag then decreases as the air density reduces, thrust is now greater than drag so the aircraft accelerates.
It is exactley the same theory as a normal, constant IAS climb in an aircraft with a flat rated engine. For the same thrust and IAS, TAS will increase with altitude"
If it is a constant ias climb (can we talk cas?) as you say, then how can drag decrease (compressibilty aside)?
Secondly, I know engines can be flat rated with respect to temperature. Can you explain this flat rating with altitude and speed? New to me.
Hawk
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straight up - like a rocket.
if u wanna go straight up, when u can not use lift created by wings u will need more thrust then weight. like rockets. u sure have seen pics of jet fighters going straight up. some of them can do the trick.
Last edited by FCS Explorer; 30th Dec 2005 at 20:41.
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I have to say I only know the basics of how a plane works. But one thing I'm still wondering about; Does the Thrust generated by a plane's engines need to be greater than the weight of the plane, for it to move, and thus take off?
The answer is YES, that is if you want to take off with brakes ON!
The static friction coeff on a locked tire on dry asphalt is Cd=1.
This means "breakaway" thrust needed is the same as the Gross weight.
Well, this is "almost" the truth since some of the acf
weight is on the nose gear which does not have brakes
Cheers,
M
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Wiz. Here's another way to consider it. Newton and F=ma.
You've said true airspeed will be increasing. Thus aircraft is accellerating. thus unbalanced force on the aircraft. And simply since the case is going straight up, the up force greater than the down force. thus Thrust greater than weight plus drag. Thus if thrust equals weight plus drag, aircraft cannot maintain same ias.
'm with Hawk on this on
Stan
You've said true airspeed will be increasing. Thus aircraft is accellerating. thus unbalanced force on the aircraft. And simply since the case is going straight up, the up force greater than the down force. thus Thrust greater than weight plus drag. Thus if thrust equals weight plus drag, aircraft cannot maintain same ias.
'm with Hawk on this on
Stan
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F=ma with F being the resultant force is the only way to consider it and says it all. Newton didn't spend all that time sitting under apple trees so that future generations could invoke witchcraft and mumbo-jumbo to explain physical systems.
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I know engines can be flat rated with respect to temperature. Can you explain this flat rating with altitude and speed?
But the engine has less air to work with, of course, so a full-rated takeoff at this altitude is naturally a bit harder on the hot section.
Somehow you can get 100,000 pounds of vertical force by only supplying 30000 pounds of horizontal force. There's gotta be a way to make perpetual motion machine out of this.
However, one fact is helping me understand:
Notice helicopters always have engines that can produce as much thrust as they weigh. If I'm wrong on this, please correct.
I'm guessing the difference is taken from the energy in the air itself. Lower energy air (turbulent) is produced and remains until heat and gravity restore said energy. Helicopters can't benefit from this in a hover.
Hey, I can guess, can't I?
However, one fact is helping me understand:
Notice helicopters always have engines that can produce as much thrust as they weigh. If I'm wrong on this, please correct.
I'm guessing the difference is taken from the energy in the air itself. Lower energy air (turbulent) is produced and remains until heat and gravity restore said energy. Helicopters can't benefit from this in a hover.
Hey, I can guess, can't I?