question regarding descent point....

[golfpapa]

Hi all
I am preparing for interviews these days even though i m not giving any but just to be prepared:)

I read that the the heavier the aircraft the earlier its descent point as compared to a lighter plane but i am not able to understand the reasoning ( from ace technical) , anyone who can shed a light on this?????:confused:

[beinghuman]

I'll give you a good reference.

http://www.pprune.org/professional-pilot-training-includes-ground-studies/412222-why-does-aircraft-descend-quicker-when-lighter.html

Hope you find it helpful.

I'll tell you how I understand it. I consider the lift formula.

L = CL 1/2 Rho V square S ( CL = AOA and Camber )

If weight increases, required lift increases.

Keeping other factors constant, increase in weight leads to increase in Velocity (Increase in groundspeed). Hence, we start out descent point sooner. Correct me if I'm wrong. Thanks.

[golfpapa]

thanks for replying

all the more confused now:sad:
1. how can rho be const. in descent?
2. the way i see it more weight means more lift required means higher AOA means lower airspeed and hence lower groundspeed??????
any clue??????:confused::confused:

tried to go thru the other thread not much help....:confused:

[beinghuman]

Alright, lets consider it this way. This will be easier to comprehend.

Are you aware of the forces involved in a descent. They are Lift, Weight, Thrust, drag and forward component of weight.

Increasing the weight, increases the forward component of weight. Hence the ROD increases and vice versa. If the ROD increases, the forward speed increases. While descent, there is a limit speed in most cases. So if we have to keep the speed under a certain limit, we have to start at an earlier point so that we can descent at a lesser ROD.

Is that helpful ? If you're finding it difficult to understand the forces in a descent, message me. I'll email you a presentation I have for descent. It graphically explains you the forces and it will be much easier to understand.

Regards

[capt. solipsist]

hopefully an even easier explanation:

think Newton's Law of Motion: Force= Mass X Acceleration or, F=MA

think of the F as the speed, M as the aicft weight, and A as the rate of descent.

to find A, divide F by M.

F should be the same for both (constant).

M is the divisor. Therefore:

a high M value (heavy acft) results in a low quotient A (low rate of descent)

conversely, a low M value (lighter acft), results in a high quotient A (high rate of descent). :ok:

[golfpapa]

thank you guys!!!!!

really appreciate all the help:D:D

[paulsalem]

All of this has a relation to airpseed too. If you didn't have to worry about Vmo or Mmo it wouldn't be an issue, as you could just descend at 500kias to make your crossing restriction.

But for instance, in the CJR-200 a "heavy" aircraft would give you 2800fpm at 290kias, a light aircraft could give you 3200fpm at 290 kias. I'm sure this is even more pronounced in a wide body.

So basically in a light aircraft at (flight idle) will be able to trade a given airspeed for a higher rate of descent than in a heavy aircraft. It's all about conservation of energy and the change from potential to kinetic.

One of the first things I learn in a new aircraft (after the type rating) is what descent rates you get at common speeds and weights. Do the math properly and you can give your pax a nice smooth fuel efficient flight idle descent without the spoilers or speed brakes. You know you've failed when you have to pull the boards out. (Unless ATC suddenly needs a higher rate of descent or your miles to touchdown changes.)

Also don't forget to take into account that a heavy aircraft takes longer to slow down than a light aircraft. Include that in your planning to give yourself enough time / distance to slow to min flap speed, so you can avoid using the boards.

[stubby1]

Can it also be explained as follows:

rate of descent = deficit of power/ wt

as deficit of power ( minimum being flight idle) being constant, heavier the ac, less the rod.;)

[golfpapa]

Vmcg cannot be less than V1...
not making sense to me considering its minimum and not maximum ground control speed...:confused::confused::confused::confused:

can somebody please shed some light on this

thanks in advance:8

[de facto]

Vmcg cannot be less than V1...

Your V1 will likely always be above VMCG unless your aircraft is very light,you have a quite contaminated runway(v1 reduction is high) and the runway you are using is quite short(runway limited),your dry v1 will be low in the first place.(minimum v1).

Now if an engine fails at a speed at vef(1sec before v1),the CAS not GS must be high enough(ie above Vmcg) to allow the pilot in a GO decision to control the aircraft with primary flight control(ie no nose wheel steering).