Load factor during a stabilized climb or descend

[manucordier]

Hi,

I read in many books that the load factor (= L/W) equals 1 in a stabilized climb or descend. However I don't agree. Could someone tell me if my reasoning is right (or wrong) ? Thanks.

So,

L = Lift
W = Weight

* Load factor = L/W

During a stabilized climb :

"a" is the angle of climb,
(= angle between the local horizontal and the trajectory of the aircraft)

* L = W cos(a)
* Thrust = Drag + W sin(a)

During a stabilzed descend :

* L = W cos(a)
* Drag = Thrust + W sin(a)

Conclusion :

In both cases: L = W cos(a)

Hance Load factor = L/W = cos(a)

However, cos(a)<1.

Therefore in a stabilized climb or descend the load factor is always smaller than 1.

Do you agree ?

Thanks.

[zondaracer]

Yes you are right, but think about this... at 10 degrees in a steady climb, load factor is .98. even at 12 degrees, it is just below .98.

also, in a vertical climb like in a fighter jet or acrobatic plane, the wings will produce 0 lift that contributes to the climb. cos 90 = 0, so load factor would be 0. the thrust handles all the load.

This is what I learned at Bristol. If I´m wrong, someone else can correct me.

[FlyingStone]

Agreed, but if you look at this equation:

L = W * cos(a)

Let's say you aren't flying an F16 or similar high-performance aircraft and you are maintaining a 5.2% climb gradient, which results in a climb angle of approximately 3°. The basic trigonometric mathematics teaches you that you can approximate values of trigonometric functions (and make many complex problems very simple) by using the following approximations:

sin(a) = a; for very small a
cos(a) = 1; for very small a (but not as small as is the case with sine)

So if you have cosine of 3 degrees, this can be taken as a very small angle, so you can just presume it's value is 1. Although if you calculate cosine of 3°, you get 0,9986, which is actually your load factor as you correctly pointed out with transforming the equation into:

L/W = LF = cos(a)

But you have to understand (I presume you are an ATPL student): ATPL syllabus covers a lot of subjects, but most of them are very similar and should be the theoretical knowledge of an average pilot. Therefore, it doesn't really matter if the book says (even though I'm pretty sure Jeppesen ATPL Manual doesn't state that) that load factor during descent is 1, since it is approximately one and you as a pilot don't need to know the precise number, because it really isn't relevant. It can be compared for example, for ATPL knowledge level it is enough to know that the pressure in the atmosphere is falling exponentially, but you don't need to know the exact formula or even how to get the formula. Or for example, it's good to know how the presence of water vapour affects air density, but it's not neccessary for a pilot to know how partial pressure of water vapour is calculated, etc.

Now: don't get me wrong, it's nice to see that anybody even takes time to actually study the subjects, but there is a lot of things covered in the ATPL syllabus and if everything would be explained with an accuracy of a physicist, you wouldn't be studying a bookset, but probably a library or two

[manucordier]

Thanks a lot guys !

[manucordier]

I made the following sketch to support my point just in case somebody would like to have a more graphic explanation :

http://cours-de-math.eu/climb-L-W.jpg

Good day,
Emmanuel.

[titinius]

When an aeroplane performs a straight steady climb with a 20% climb gradient, the load
factor is equal to:
0.83
1
0.98
1.02

load factor = L/W

L = W . cos angle
angle = 20° = 0,2
L/W = cos 0,2
L/W = 0,9396 ...

-> there is no right answer ?? (0,98 is marked as a right one)

[manucordier]

titinius,

A 20% climb gradient means that

TAN(angle of climb) = 20/100 = 0,2

So ARCTAN(0,2) = angle of climb = 11,31 degrees.

Now, Load Factor = L/W = COS(angle of climb) = COS(11,31) = 0,98.

So LF = 0,98 in a 20% climb gradient.