Header1200x385

× Welcome to the CPL Aerodynamics question and answer forum. Please feel free to post your questions but more importantly also suggest answers for your forum colleagues. Bob himself or one of the other tutors will get to your question as soon as we can.

Forces acting

  • lcanning
  • Topic Author
  • Away
  • Posts: 17
  • Thank you received: 0

lcanning created the topic: Forces acting

G'day team. Hope all is well.

Yesterday I got a question in my exam asking me "what are the forces equal to in straight and level flight".

Looking through the book I have found that During the
Decent - Lift is less than weight & Thrust in less than drag.
Climb - Lift is less than weight & thrust is greater than drag

I have not been able to find anything in the book about straight and level flight all though it may be obvious to say all components would be equal but I think I may have second guess myself and chosen wrong

2nd Question is - When a Aircraft is in decent or climb and they are neither accelerating or decelerating is this considered Equilibrium

Thanks Team,
#1

Please Log in or Create an account to join the conversation.

  • John.Heddles
  • Offline
  • ATPL/consulting aero engineer
  • Posts: 874
  • Thank you received: 107

John.Heddles replied the topic: Forces acting

It can get a bit confusing.

The way we approach figuring force balances is to look at all the forces which are acting on the object under consideration (in our case, an aeroplane). Conventionally, we resolve (buzzword meaning "break them down into") the various forces in orthogonal (a fancy buzzword for "at right angles") directions which allows us to make easier mathematical sense of what is going on.

If the aeroplane is NOT accelerating in any direction then there can not be any unbalanced forces at play, that is to say, each and every force is balanced by other forces or parts of other forces. We refer to this sort of arrangement as the forces "being in equilibrium" which just means that there is no imbalance in, or residual bits of, the forces causing changes in what is going on. This means that, if the forces are in equilibrium, then there can not be any accelerations, ie the aeroplane will be moving at a constant speed, either level, or in climb or descent. Generally, we also like to balance the torques (or moments) so that there is no residual rotation going on. However, we don't tend to get too excited about this aspect of equilibrium in pilot training.

Conversely, if the aircraft IS accelerating then there has to be some imbalance in the forces at play to cause the acceleration.

Strictly we can't look, say, at lift and weight or thrust and drag in isolation as they act together to produce an end result.

When we suggest that "lift is less than weight" we are looking, in isolation, at those two forces when we should always be looking at the whole picture, not just bits and pieces of it.

Looking at the question you pose "what are the forces equal to in straight and level flight", this really doesn't mean all that much from the point of the physics at play. However, we are just trying to simplify things a bit to make it easier for the new pilot to get a handle on what is going on. So, not quite correctly, but usefully, we think in terms of the lift force (ie the total of the various forces collectively helping keep the aircraft up in the air and at a constant altitude, balancing all the forces trying to drag the aircraft down to the ground (which is, mostly, the weight force) as being equal. Hence we might talk about "lift" being equal to "weight". This is a bit rubbery as there is a collection of forces at play which, collectively, achieve the final result. However, if we drill down into the story's details, things would be quite a bit more complicated for no real benefit to the new pilot in terms of his or her understanding of what is going on.

If you need me to do so, I can expand the above story as might be required to help you get your head around what is going on. However, keep in mind that the system is not trying to make physicists or engineers of pilots but rather, trying to get them to a basic level of understanding sufficient to have an idea of what is going on. The important thing is to have enough knowledge to help the stick and rudder learning, not design aeroplanes.

Engineering specialist in aircraft performance and weight control.
#2

Please Log in or Create an account to join the conversation.

  • lcanning
  • Topic Author
  • Away
  • Posts: 17
  • Thank you received: 0

lcanning replied the topic: Forces acting

John this is great.

So just to clarify - In level flight with any acceleration or decelerating. Lift is = to weight & Thrust is = to drag

and for my second question aircraft in a climb lift is less than weight & Thrust is more than drag BUT this is still considered to be in a state of equilibrium as long as there is no constant change in force like acceleration or deceleration
#3

Please Log in or Create an account to join the conversation.

  • John.Heddles
  • Offline
  • ATPL/consulting aero engineer
  • Posts: 874
  • Thank you received: 107

John.Heddles replied the topic: Forces acting

Note the correction -

In level flight withOUT any acceleration or decelerating. Lift is = to weight & Thrust is = to drag

That is a reasonable approximation to what is going on and quite fine for pilot training work.

aircraft in a climb lift is less than weight & Thrust is more than drag

Yes, however, the important thing is that the overall ("resultant") of the various force vectors (buzzword meaning that the item has both a magnitude, or size, and direction, or sense) are such that the total of the bits pointing up equals the total of the bits pointing down and the total of the bits pointing forward equals the total of the bits pointing backwards. The statement refers to the individual vectors which relate to wing lift, weight, thrust and drag. The overall picture, though, requires that we look at all the vectors in conjunction. Really the usual pilot statement is not a great deal of use to anything but it is what the examiner wants to see you put down in the exam.

BUT this is still considered to be in a state of equilibrium as long as there is no constant change in force like acceleration or deceleration

In this sort of work, "equilibrium" means that everything nicely balances out and there is no residual acceleration - the aeroplane keeps going along quite nicely at a constant speed, ROC, ROD or whatever is of interest to us.

Engineering specialist in aircraft performance and weight control.
#4
The following user(s) said Thank You: lcanning

Please Log in or Create an account to join the conversation.

  • lcanning
  • Topic Author
  • Away
  • Posts: 17
  • Thank you received: 0

lcanning replied the topic: Forces acting

Thank you John for your help. This is great
#5

Please Log in or Create an account to join the conversation.

  • Posts: 1182
  • Thank you received: 140

Stuart Tait replied the topic: Forces acting

Louis are you able to login to the onlineschool? online.bobtait.com.au/my/ I'm getting a bounce back on your email, If not contact

Maree in the office This email address is being protected from spambots. You need JavaScript enabled to view it. to sort it out.
#6

Please Log in or Create an account to join the conversation.

Time to create page: 0.302 seconds