Load factor = Lift / Weight
If we take it to the limit when an aeroplane can generate more thrust than its weight then can it can enter a stable vertical climb with zero lift?
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PDR
I may have misread your post if you were saying the the Vertcal Components of Thrust and Drag are equal rather than Thrust and Drag are equal.
Of course the relationship between IAS and TAS is a function of air density. But Density is not a Force and plays no part in Newton’s Laws. If we are going to use vector mechanics we need to stick to the Rule of Law.
Custard
You have the First Law right. But in a 70 kt IAS climb the aicraft is not in a steady state but is, indeed, accelerating.
IAS is a red herring. Our ‘frame of reference’ for Newton is the Earth and the appropriate velocity is TAS not IAS.
To an observer on the ground, as you climb at 70 kt IAS you will be seen to accelerate. TAS will increase.
As you maintain a, commendably steady 70 kt IAS climb, to an observer in the aircraft looking out of the window, you will be seen to accelerate in relation to the ground. TAS will increase.
This is is real life. And, according to Newton’s Second Law the forces on the aircraft must be unbalanced.
In our vector diagram for the climb Thrust must be greater than Drag.
Or, if you like, Drag must be less than Thrust
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Of course, in vertical flight, while this contraption is an still an Aircraft it is not an Aeroplane/Airplane. Not producing lift by accelerating a fluid over an inclined Plane.
Plane (AKA wing) as in Bi-plane, Mono-plane, Tri-plane.
I don't know, there are six inclined planes at the front of this thing, all of them producing lift by accelerating a fluid medium.... 
Oh all right, the EASA definition agrees with you, I'll get my coat then...
‘Aeroplane’ means an engine-driven fixed-wing aircraft heavier than air, that is supported in flight by the dynamic reaction of the air against its wings.
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You are right. Just like a Rotary Aircraft those propeller blades (Inclined planes) are producing a force at right angels to the Relative Airflow over the blades: Lift by standard definition.
Indeed in any propeller driven aircraft the propeller produces a force at Rt Angles to the Relative Airflow over the propeller blades. But I will continue, for clarity at least, to call this force thrust rather than lift.
For me Rotary Aircraft are Rotary Aircraft not Aeroplanes. But I don’t mind what the general vernacular is.
A picture tells a thousand words, so three pictures tells a short novel! I've had time to do some sketches to show what I'm saying:
I'm slightly mystified as to why Dutystude introduced IAS/TAS issues into this - unless I'm missing something it's irrelevant to the subject. I can't think of any rule of law that I've infringed. If I have please point me at it!
PDR
I'm slightly mystified as to why Dutystude introduced IAS/TAS issues into this - unless I'm missing something it's irrelevant to the subject. I can't think of any rule of law that I've infringed. If I have please point me at it!
PDR
Last edited by PDR1; 5th Nov 2018 at 12:30. Reason: The keyboard typed "thing" when I typed "think". My gamekeeper is giving the keyboard a sound correctional horsewhipping
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Hello PDR,
First, if I might steal a line from Shakespeare’s Mark Anthony: “I speak not to disprove what PDR has spoke but to say what I do know.” I am enjoying the intellectual exchange and not looking for a fight.
What I know from many years of observation is this: if I fly my ac at sea level at 70 KIAS my TAS will be 70KTAS. If I apply full power and maintain 70KIAS in the climb when I pass 10000 ft (as if) my TAS will be 80KTAS or thereabouts.
This is what I know. The why bit has been of no practical use to me as a pilot but, perforce, I have been a flying instructor a number of times and been required to offer a plausible explanation to the moe inquisitive Tyros that I have had the pleasure to teach.
You ask why I bang on about IAS/TAS.
Well, to begin with Indicated Air Speed is not actually a Speed except at sea level in ISA conditions. Better to remember that what you see on the ASI is Dynamic Pressure delivered via the pitot tube. The ASI could equally be gauged in inches of Mercury or Hecto Pascals. But, I’m quite pleased that they chose MPH/Kts.
Dynamic Pressure, in simple terms, is composed of the speed of the air and the density of the air. This speed is indeed a speed. It is the speed of the air in our ‘frame of reference’ – Planet Earth. It is the speed that the ac is moving in the surrounding air and, by convention, we call it True Air Speed (TAS). It is the speed used in Kinetic Energy, Momentum, acceleration and Newton’s Laws.
So for we pilots IAS (dynamic pressure) depends on TAS (speed through the air) and air density. If the density reduces (unavoidable in a climb) IAS will reduce.
But it doesn’t, we know that. But what we forget (because, for most it is of no practical value) while we maintain IAS, TAS is increasing (The ac is accelerating ) to balance the drop in air density.
The ac is not accelerating under the influence of air density. In accordance with Newton’s second Law the ac can only accelerate if acted upon by an external force – density is not a force.
As to vector diagrams:
In level balanced un-accelerated flight, Forces on the ac are not overcoming weight (gravity) they are balancing weight.
In a constant IAS climb the Forces on the ac are not balancing weight, they are overcoming weight, otherwise the ac would not move away from the centre of the Earth – climb that is. And, the Forces on the ac are accelerating the ac in the surrounding air – TAS increasing.
Some force is raising the ac in relation to the Earth’s surface; some force is accelerating the ac. If, a chosen vector diagram does not show this then I guess, as many suspect, it is just FM - Flipping Magic.
Which, when I have been in the mood, I have, sometimes, given as an explanation to many of the mysteries of flight.
First, if I might steal a line from Shakespeare’s Mark Anthony: “I speak not to disprove what PDR has spoke but to say what I do know.” I am enjoying the intellectual exchange and not looking for a fight.
What I know from many years of observation is this: if I fly my ac at sea level at 70 KIAS my TAS will be 70KTAS. If I apply full power and maintain 70KIAS in the climb when I pass 10000 ft (as if) my TAS will be 80KTAS or thereabouts.
This is what I know. The why bit has been of no practical use to me as a pilot but, perforce, I have been a flying instructor a number of times and been required to offer a plausible explanation to the moe inquisitive Tyros that I have had the pleasure to teach.
You ask why I bang on about IAS/TAS.
Well, to begin with Indicated Air Speed is not actually a Speed except at sea level in ISA conditions. Better to remember that what you see on the ASI is Dynamic Pressure delivered via the pitot tube. The ASI could equally be gauged in inches of Mercury or Hecto Pascals. But, I’m quite pleased that they chose MPH/Kts.
Dynamic Pressure, in simple terms, is composed of the speed of the air and the density of the air. This speed is indeed a speed. It is the speed of the air in our ‘frame of reference’ – Planet Earth. It is the speed that the ac is moving in the surrounding air and, by convention, we call it True Air Speed (TAS). It is the speed used in Kinetic Energy, Momentum, acceleration and Newton’s Laws.
So for we pilots IAS (dynamic pressure) depends on TAS (speed through the air) and air density. If the density reduces (unavoidable in a climb) IAS will reduce.
But it doesn’t, we know that. But what we forget (because, for most it is of no practical value) while we maintain IAS, TAS is increasing (The ac is accelerating ) to balance the drop in air density.
The ac is not accelerating under the influence of air density. In accordance with Newton’s second Law the ac can only accelerate if acted upon by an external force – density is not a force.
As to vector diagrams:
In level balanced un-accelerated flight, Forces on the ac are not overcoming weight (gravity) they are balancing weight.
In a constant IAS climb the Forces on the ac are not balancing weight, they are overcoming weight, otherwise the ac would not move away from the centre of the Earth – climb that is. And, the Forces on the ac are accelerating the ac in the surrounding air – TAS increasing.
Some force is raising the ac in relation to the Earth’s surface; some force is accelerating the ac. If, a chosen vector diagram does not show this then I guess, as many suspect, it is just FM - Flipping Magic.
Which, when I have been in the mood, I have, sometimes, given as an explanation to many of the mysteries of flight.
there must be an initial acceleration when the aircraft enters a climb from S&L.
To cause that change requires unbalanced forces. It could be done by increasing power or changing angle of attack (or a combination).
This unbalance changes the velocity of the aircraft (ie causes an acceleration) from a level direction to one angled upwards.
In the 'steady' climb that (hopefully) results* after this change all the forces will again end up balanced, with the magnitude of the lift force being less than weight, hence my original statement, lift can be less than weight in some modes of flight).
*ok, in real life we cannot sustain a steady climb forever but for small climbs of a few thousand feet, we should be able to approximate reasonably closely a steady climb.
To cause that change requires unbalanced forces. It could be done by increasing power or changing angle of attack (or a combination).
This unbalance changes the velocity of the aircraft (ie causes an acceleration) from a level direction to one angled upwards.
In the 'steady' climb that (hopefully) results* after this change all the forces will again end up balanced, with the magnitude of the lift force being less than weight, hence my original statement, lift can be less than weight in some modes of flight).
*ok, in real life we cannot sustain a steady climb forever but for small climbs of a few thousand feet, we should be able to approximate reasonably closely a steady climb.
PDR1
In your second diagram the aircraft is climbing at constant speed and constant roc, so the forces must be in equilibrium. You show the thrust and drag as being in equilibrium, but we need to look a bit more closely at lift and weight
Let us start by sliding the lift arrow down and to the left until it's point touches the tail of the weight arrow. We now have two sides of a triangle of forces. For these forces to be in equilibrium, there must be a third force arrow joining the point of the weight arrow to the tail of the lift arrow.
You have shown lift being greater than weight, which means that lift is the hypotenuse and the third force must be at right angles to the weight. This means that we must have a horizontal force pointing to the left. What is this force?
If you are wrong and lift is less than weight, then the weight is the hypotenuse. In this case the third force arrow is acting up the flight path and meets the lift arrow at right angles. In conventional texts this is the excess thrust.
Now let's see where your arguments lead us if we have a very high-powered aircraft such as a modern fighter jet. If your diagram is correct it means that as we gradually increase the climb angle, the lift force must increase. When the climb angle is 90 degrees in a vertical climb, the lift force must be infinite. But no wing can produce infinite lift, so logically if your diagram is correct, no aircraft could ever achieve a vertical climb.
So, is it impossible to achieve a vertical climb, or are you mistaken in your arguments?
In your second diagram the aircraft is climbing at constant speed and constant roc, so the forces must be in equilibrium. You show the thrust and drag as being in equilibrium, but we need to look a bit more closely at lift and weight
Let us start by sliding the lift arrow down and to the left until it's point touches the tail of the weight arrow. We now have two sides of a triangle of forces. For these forces to be in equilibrium, there must be a third force arrow joining the point of the weight arrow to the tail of the lift arrow.
You have shown lift being greater than weight, which means that lift is the hypotenuse and the third force must be at right angles to the weight. This means that we must have a horizontal force pointing to the left. What is this force?
If you are wrong and lift is less than weight, then the weight is the hypotenuse. In this case the third force arrow is acting up the flight path and meets the lift arrow at right angles. In conventional texts this is the excess thrust.
Now let's see where your arguments lead us if we have a very high-powered aircraft such as a modern fighter jet. If your diagram is correct it means that as we gradually increase the climb angle, the lift force must increase. When the climb angle is 90 degrees in a vertical climb, the lift force must be infinite. But no wing can produce infinite lift, so logically if your diagram is correct, no aircraft could ever achieve a vertical climb.
So, is it impossible to achieve a vertical climb, or are you mistaken in your arguments?
Last edited by keith williams; 6th Nov 2018 at 06:29.
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In this diagram weight has been resolved into 2 component vectors at right angles; the long side equal and opposite to the Lift vector. You will notice that in this vector triangle Weight is the hypotenuse of the triangle and therefore, by observation, is longer (so greater) than the vector that is equal and opposite to the Lit vector.
QED: In a climb Lift is less than weight.
If so we then only need to ensure lift and weight cancel each other for a steady motion.
W is shown with a rearwards component to the flightpath direction - is that correct?
L is also shown acting with a rearwards component to the flightpath - is that correct?
In that case don't we have an unbalanced set of forces with a net force acting rearwards to the flightpath?
This then cannot be a steady motion could it? - your diagram is showing a situation where the aircraft would be de-accelerating.
ie diagram 2 shows an unbalanced set of forces where there would be an acceleration slowing the aircraft, not a steady, unaccelerated climb.
I have a sneaking suspicion you know this and are wanting to see how far you can pull the wool
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I don’t think these vector diagrams are correct when they have the weight always pointing straight down. That situation is only possible when the weight is stationary.
if the aircraft is moving the weight is acting against that movement, along with the aerodynamic drag. (Modified by gravity of course)
so the W vector should be angled rearward by the addition of an inertia vector.
Taking all these vectors as they are usually displayed, as if the aircraft were stationary, Then remove the lift, thrust and drag, but leave the fact that the aircraft is moving, such as would happen if it left the atmosphere, then the aircraft would not just fall vertically down the W vector, it would continue forward/downward on a curved path as the inertia vector decayed to zero and replaced by W.
if the aircraft is moving the weight is acting against that movement, along with the aerodynamic drag. (Modified by gravity of course)
so the W vector should be angled rearward by the addition of an inertia vector.
Taking all these vectors as they are usually displayed, as if the aircraft were stationary, Then remove the lift, thrust and drag, but leave the fact that the aircraft is moving, such as would happen if it left the atmosphere, then the aircraft would not just fall vertically down the W vector, it would continue forward/downward on a curved path as the inertia vector decayed to zero and replaced by W.
No excuses - saw something which made me jump to a particular conclusion but completely failed to do a proper analysis before spouting off. Please accept my apologies.
PDR
In the course of my life I have often had to eat my words, and I must confess that I have always found it a wholesome diet.
PDR and Jonkster, you are in highly esteemed company!
Before reading this discussion I had rather assumed that the 1G stalling speed of an aircraft is the same (when it is in steady, straight, unaccelerated flight) whether it be level, climbing or descending. I now realise that the 1G stalling speed is at a maximum in level flight and reduces if the aircraft is either climbing or descending because the wing loading reduces in the climb or descent.
