View Full Version : Turn rate, aerodynamic question

Double Back

28th Jun 2011, 21:21

I wonder what the relationship is with Bankangle, Speed (IAS) and turn rate(heading change).

Clarified:

A plane flies with 10 degrees of bank and starts to reduce its speed (IAS)

The radius of the turn will decrease, that's obvious, but I also think the heading change(per minute) will increase.

Flying an airliner at cruise speed makes a heading change take for ages with let's say 10 degrees of bank.

While on final the heading changes faster with the same 10 degrees of bank.

How this can be explained or am I completely wrong?

Richard

fantom

28th Jun 2011, 21:58

I suggest you start here:

http://www.amazon.com/Mechanics-Flight-10th-C-Kermode/dp/0582237408

fantom

28th Jun 2011, 22:00

Ah, that might be pulled for whatever reason.

Start again:

I suggest you start with Kermode's 'Mechanics of Flight'.

Checkboard

28th Jun 2011, 22:04

Or you could just play around with a turn calculator:

Aircraft Turn Information Calculator (http://www.csgnetwork.com/aircraftturninfocalc.html)

Mad (Flt) Scientist

28th Jun 2011, 22:09

Assuming the aircraft is maintaining a level turn, the bank angle can be related to the centripetal force with the assumption that there is no significant sideforce being generated and that the (wing) lift is the dominant force.

So assuming a vertical (earth axis) force of W (=mg) the aircraft weight is generated, and is the vertical component of the total Lift L, then the sideforce SF in earth axis terms is simply the third side of the triangle of forces comprising W, L and SF. The bank angle B is known and is the angle between the W and L force vectors, so SF is the opposite side to B and L is the hypoteneuse, so simple trig says that SF=WtanB (so with bank angle 0, tanB=0 and so SF=0, as we'd expect, and we get nonsense answers as B approaches 90, because we can't sustain the constraint for level flight with our assumptions in a 90 bank attitude.)

So we know now that SF=WtanB = mgtanB.

We also know from simple mechanics that the centripetal force is a function of the mass of the body,m, its speed,V, its angular rate, a, and the radius of the turn, R, as follows:

Force = m * V * V /R = m * a * a * R = m * V * a

(usually the symbol used for angular rate is omega)

if we substitute in the SF term for the Force, and cancel out the common mass term, m, we get:

g * tan B = V * V / R = a * a * R = a * V

So, for a fixed bank angle B, the left side is constant. Therefore each of the other equivalent expressions must be constant. So:

if we decreased speed (V) in a constant bank angle turn, then the radius of turn must decreased (to keep V * V / R constant) and the angular rate (heading change rate) must increase (to keep a * V constant)

ChristiaanJ

28th Jun 2011, 22:11

Another classic...

http://www.amazon.com/Aerodynamics-Naval-Aviators-FAA-Handbooks/dp/156027140X/ref=pd_sim_b_1

And I do hope the mods will understand that references to a particular book will often have to pass through a reference to a bookseller.

CJ

fantom

28th Jun 2011, 22:25

Mad FS:

Go and lie down in a dark room; we can't keep up.

Capn Bloggs

29th Jun 2011, 02:02

Or you could just play around with a turn calculator:

Aircraft Turn Information Calculator

They got everything except the °/sec turn rate! :p

Double Back

29th Jun 2011, 14:50

Hello Mad Scientist

Like many other pilots I hate formulae..

As You seem to know how the mathematics in my question work out, can You explain it in normal language, plus maybe a small drawing if possible?

I need to explain it too to some guys, at the moment my speech would be weak....

FlightPathOBN

29th Jun 2011, 16:00

DoubleBack,

Here are the turn radius calculators from 8260.52...

Formula 1-2 assumed a fixed bank angle of 18 degrees, so similar to your question. As you can see, changing the speed, reduces the turn radius, as given that a turn is a rate of change, the rate of change increases.

http://operationsbasednavigation.com/wordpress/wp-content/uploads/2011/04/Formula1-2.jpg

Formula 1-3 allows one to vary the bank angle...

http://operationsbasednavigation.com/wordpress/wp-content/uploads/2011/04/Formula1-3.jpg

Good Luck!

Dont Hang Up

29th Jun 2011, 16:58

Ok skip the maths. There is a simpler way to think about it.

If an aircraft travelling at 300 Knots wants to reverse direction in one minute (6 degrees per second) then this is a velocity change of 600 Knots.

It must obviously pull a higher 'g' than an aircraft doing the same turn at only 100 Knots (change of velocity only 200 Knots).

'g' relates to bank angle with no dependency on speed (for example 60 degree of bank is 2g nomatter how fast or slow you are going). Therefore the faster aircraft must pull a higher bank angle to achieve the same turn rate.

FlightPathOBN

29th Jun 2011, 17:22

sorry, but I need to hang up on that one.....:cool:

rennaps

30th Jun 2011, 13:00

No one has mentioned altitude which will also have a bearing on TAS

john_tullamarine

30th Jun 2011, 13:32

.. only if you fancy making a simple equation drearily complex ... and only for the sake of changing TAS to IAS ?

STOLskunkworks

30th Jun 2011, 13:51

explained by inertia.

If you want to explain it just do turns in you car at different speeds in the parking lot but try to keep the same feeling in terms sideways g's.

you could also use the bank of a turning bike if that helps

cwatters

30th Jun 2011, 14:45

A plane flies with 10 degrees of bank...

A plane flying with 10 degrees of bank need need not be turning at all.

YouTube - ‪8 Point Roll - http://www.wild-wings.co.uk‬‏

The short answer is that without imposing further constraints there is no _fixed_ relationship between Bankangle, Speed (IAS) and turn rate(heading change). It can be changed to give different outcomes.

Flying an airliner at cruise speed makes a heading change take for ages with let's say 10 degrees of bank. While on final the heading changes faster with the same 10 degrees of bank.

For comfort you want any g forces acting on passengers to be towards their feet not sideways. In a turn they experience two forces: regular gravity (towards the ground) and centrifugal force (horizontally due to the turn). The plane normally banks so that the combination of these two forces is towards their feet.

Centrifugal force depends on both the radius of the turn and the speed of the plane. On approach the plane is going slower so the turn can be tighter for any given for any given bank angle.

Double Back

30th Jun 2011, 15:17

Cwatters

OK, nice SIM demo!

Of course I am interested in "straight" (coordinated)turns, not any fancy rudder adjusted roll manoeuvres.

There IS a relationship for the parameters I based my question on. I do got a lead now I just got from a befriended aeronautical engineer. The above mentioned formulae are beyond my comprehension.

Need some time to translate it, maybe make a drawing.

Quote:

Centrifugal force depends on both the radius of the turn and the speed of the plane. On approach the plane is going slower so the turn can be tighter for any given for any given bank angle.

In my opinion a misunderstanding of flight physics.

At a certain speed a given bank angle will induce a centrifugal force (some might dub it as centripetal) and this will "draw" the imaginary cylinderwall the plane follows. The radius.

Not the other way around.

Planes normally fly bank angles and adjust that to get the correct (ground referenced)radius to roll out on the extended C/L from baseleg for instance. As a pilot You fly bank, not radius.

The last airliner I flew however was the 747-400 who had some kind of "prediction mode" for making a turn by showing a self adjusting and curving arrow that You could more or less use to HELP you to get yourself lined up from baseleg.

The smart (calculated) arrow was based on groundspeed and ground radius.

Mad (Flt) Scientist

30th Jun 2011, 21:08

Like many other pilots I hate formulae..

As You seem to know how the mathematics in my question work out, can You explain it in normal language, plus maybe a small drawing if possible?

OK, I'll try.

There's two parts to the answer. The first is the fact that the cornering force is the same for a fixed bank angle, regardless of speed. For this one, you'll need to draw a little diagram (I dont have an easy way to attach a sketch).

On a sheet of ideally lined or squared paper, draw a little aeroplane viewed from behind, at some fairly large bank angle - 30 degrees or so will be enough to show it. draw an arrow pointing down, and label it "weight". draw a dotted line, at right angle to the wings, pointing "up" as seen by a pilot (so sloping on the page) - that's the direction the lift acts in. draw another arrow, equal and opposite to the weight, pointing up. That's the part of the lift which will be opposing the weight and keeping us in level flight.

You can see that the lift isn't acting in the direction we want. So now draw an arrow along the dotted line, long enough so that it stops at the same height as the "anti-weight" force line. That's how much lift you need, so that the portion, or component, which is available to combat the weight is the right size for level flight.

Note now that there is a gap between the "anti-weight" arrow and the lift arrow. Draw an arrow from the tip of the "anti-weight" arrow to the tip of the lift arrow. This is the part, or component, of the lift force which ISNT fighting the weight - and it's pointing sideways. It's this force - which is there because we are banked - which will cause us to turn.

Now, you've drawn all of that without using a number once. So the lengths of the lines are just dependent on the shape of the triangle they form, and are fixed relative to each other. So for a given weight, and a given bank angle, the length of the cornering force line must be the same. So the cornering force depends only on the bank angle and the weight. Speed doesn't matter.

The second part that is important is the statement that the cornering force is given by that list of equations:

F=m*v*v/R or = m*v*a or = m*a*a*R

where v is the speed, R is the radius of turn and a is the anguler velocity or rate of turn. The middle one is the one we'll work with, since v, a and R are related to each other by v=a*R (that can be explained later on)

The common formulation of Newton's second law as an equation is F=ma. But another way talking about Newton's laws is to say that a body has a constant and fixed momentum, and that it takes a force acting on the body to change the momentum, and that the rate of change of the momentum depends on the amount of force applied.

Momentum is the combination of the mass of something and its velocity. Velocity means not just the amount or magnitude of the speed, but also the direction. So if an object (an aircraft) has a fixed mass and a fixed speed, but a changing direction, then it has a change in momentum, and that means there is a force acting on it.

It turns out that we can relate the force and the momentum change directly, so that

Force = momentum change per second.

If we break momentum down into mass times speed "times" direction (you'll have to trust me on the "times", it does make sense) then rate of change of momentum could be:

mass per second times speed time direction

OR

mass times speed per second times direction

OR

mass times speed times direction per second

with the bits in bold as the changing quantity (keeping it simple with only one thing changing at a time)

usually when the direction isn't changing we leave it out of the equations and just use the values (or 'scalars' to use the mathematical term).

So the first one becomes:

Force = mass per second times speed. That is, in fact, the "rocket equation" in a simplified form, because rockets expel mass to generate force.

The second one then is:

Force = mass times velocity per second. Well, velocity per second is acceleration. This is our old friend, force=mass times acceleration, F=ma, the standard Newtonian equation.

And the third one?

Force = mass times velocity times angular rate - the F=m*v*a we used before.

Hopefully that explanation helps explain the F=WtanB and F=m*v*a equations I used before. From those two you can then see why the terms have the relationship they do in a banked turn.

Double Back

3rd Jul 2011, 13:19

Based on remarks from the "mad" scientist, who btw I do not consider so mad at all..., and a befriended modelflying aeronautical engineer, now it looks I have understood the matter, I made a small drawing which should clarify it all. That even pilots can apprehend! :)

Sorry for the normally drawn pic, photographed, saved on PC and then to an internet site.

Distances are not to scale.

A really important given, like MS also says, is that the sideways force for a given bank angle ALWAYS remains the same, regardless of speed, I never apprehended that obviously completely, but come to think of it, it is absolutely logical.

The effect of that force, however, varies with speed.

"In all: a constant force on a constant mass results in a constant displacement in a given time span"

Everybody reading here knows those drawings that show how Lift with bank angle is broken down in a vertical and horizontal component. No need for that one.

For getting the problem, imagine a turn is broken down into 1 sec straight parts.

https://lh4.googleusercontent.com/-2VJJZd2ols8/ThBVP9No7FI/AAAAAAAADpM/qVrn7jksn7o/s640/IMG_0524.JPG

Suppose we have a 10 degree bank and a speed of 300 meters/sec, the sideways force might be around 2 mps.

The top example indicating the speed with a vector to A

AB is the offset created by the sideways force.

The final path of the plane will be to B with a track change of around 0.4 degrees.

Now the same plane at 100 mps and 10 degree bank with the vector stretching only until C.

CD is again the same 2mps offset.

It is very clear the resulting track to D is at a greater angle compared to the high speed version, like some 1.2 degrees.

The only thing needed is to accept that the sideways force also results in an aerodynamic force (the directional stability) that drives the nose to the same heading as the changed ground track, completing the actual heading change which can be seen outside from a moving horizon or on the compass/directional navigation systems.

Richard