Jbmitt From United States of America, joined Jan 2002, 588 posts, RR: 2 Posted (12 years 1 week 4 days 18 hours ago) and read 10846 times:

Hi guys/gals,

For my calculus class I'm working on a project, and one of the questions is regarding 'where should a pilot start descent'.

I came here hoping that somebody might be able to assist me with the mathematics that goes along with the problem. In addition to solving the problem I need to explain why.

The question is as follows:

An approach path for an aircraft landing must satisfy the following conditions.

given y=P(x)

(i) The cruising altitude is "h" when descent starts at a horizontal distance"l" from touchdown at the origin.
(ii) The pilot must maintain a constant horizontal speed "v" throughout descent.
(iii) The absolute value of the vertical acceleration should not exceed a constant "k" (which is much less than the acceleration due to gravity).

1. Find a cubic polynomial P(x)=ax*3 + bx*2 + cx + d (*3 and *2 are exponents, ax to the third, and bx squared but I have no idea how to type those) that satisfies condition (i) by imposing suitable conditions on P(x) and P'(x) at the start of descent and at touchdown.

2. Use conditions (ii) and (iii) to show that 6hv*2/l*2
3. Suppose that an airline decides not to allow vertical acceleration of a plane to exceed k=860 mi/h*2 (*2 is an exponent, in this case squared. If the crusiing altitude of a plane is 35,000 ft and the speed is 300 mi/h, how far away from the airport should the pilot start descent?

FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 1, posted (12 years 1 week 4 days 16 hours ago) and read 10699 times:

y would be the altitude and x the distance from the touch down point (aimpoint, see below) then?

No restraint on the maximum vertical speed [P'(x)] or airspeed? Nor on the glide path angle at touchdown? Hmmm... so you'd basically accelerate downwards at k units of acceleration until you hit the ground for the shortest approach?

(Btw, you'd write it P(x)=a*x^3 + b*x^2 + c*x + d )

Common sense would give you a descent rate just prior to the flare (to avoid having to fit the flare into the equation, assume (x,y)=(0,0) to be the aimpoint rather than the touchdown point and that the flare does not have to be described) giving you a three degree glideslope at speed v.

P'(0)/v = tan 3*

Or, if you want to make it harder, P'(x)/v = tan 3* when P(x) = 50 feet or so...

P(0) = 0

P(l) = h

P'(l) = 0

-k < P"(x) < k for 0 < x < l

Bleh, this already stinks of things I've tried hard to forget! And when I try hard, I often succeed...

Cheers,
Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Timz From United States of America, joined Sep 1999, 7210 posts, RR: 7
Reply 3, posted (12 years 1 week 4 days 14 hours ago) and read 10642 times:

Yeah, he left something out there.

I guess this has nothing to do with gravity or potential energy vs speed etc. It's given that the horizontal component of speed will be constant, and we don't care how the pilot arranges that. (Also, the earth is flat.)

The problem sets a limit on vertical acceleration (either positive or negative)-- so that limits the curvature of the plane's path (in side view) both at the top and the bottom of the descent. We are to assume the plane is flying level at the bottom (at the end of the runway) and at the top-- so P'(x) is zero at the bottom and at the top. And we're specifying P''(x) at the bottom, and I guess it will be the same absolute value at the top.

And at top of descent P(x) = h and at the bottom P(x) = 0. Sounds like enough to determine a, b, c and d.

FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 4, posted (12 years 1 week 4 days 4 hours ago) and read 10479 times:

Hmm... "(iii) The absolute value of the vertical acceleration should not exceed a constant "k" (which is much less than the acceleration due to gravity)."

Hm, the absolute acceleration? Not the absolute acceleration outside of that used to counter gravity? This could be taken to mean that we will not be alloved to have an upwards acceleration of, say, 9.82 m/s^2 or so?

A very short approach it will be then!

Timz,
if we set P'(0) = 0 and no condition on the glideslope... chances are we'll end up with a very shallow approach indeed. Depends on how real we want to keep things, I guess?

Sitting here muddying the waters over my morning coffee.

Cheers,
Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

Timz From United States of America, joined Sep 1999, 7210 posts, RR: 7
Reply 5, posted (12 years 1 week 3 days 14 hours ago) and read 10391 times:

There is no "acceleration used to counter gravity", in the real world or in the problem. I think the problem writer intends for us to forget about gravity, so we might as well do so.

The problem asks for a cubic, so P''(x) is assumed to be a straight line. The magnitude of the downward acceleration at the very top of descent equals the magnitude of the upward acceleration at the end of the runway, and that magnitude is specified.

It is your project, so I didn't want to do more, but hopefully you can work it out from here. Note that number 1 deals only with the vertical direction, and you should incorporate the constant x-velocity constraint and the acceleration (k) constraint for the later parts.

Make sure you cite the work - say you had help from airliners.net and maybe give the address of the graphic so you don't get in trouble

TimT From United States of America, joined Jun 2001, 168 posts, RR: 0
Reply 9, posted (11 years 12 months 3 days 6 hours ago) and read 10018 times:

Since I'm somewhat simple minded, I did a little looking and found AOPA's Rod Machado had written a column for the magazine that addressed this very subject. His answer:

To find the distance from the airport that you need to start descent using 500fpm as a base. Multiply your groundspeed (in miles per minute)times 2, then multiply that by by the amount of altitude (in thousands of feet) you need to lose. His example: You're descending at 180kts, and need to lose 9000 feet. 180kts=3nm/min. Thus 2 x 3=6 6 x 9 =54
Begin descent at 54nm from airport..

FredT From United Kingdom, joined Feb 2002, 2185 posts, RR: 26
Reply 10, posted (11 years 12 months 3 days 3 hours ago) and read 9997 times:

Altitude loss in thousands of feet times three is often good too. Wanna come down from FL170 to 8000'? (17-8) * 3 = 9*3 = 27. That's a constant descent angle.

The AOPA method is based on a constant 500 fpm descent rate. The big iron tends to come down faster, in which case the method isn't valid.

Even if you up the fpm figure to what is typical for whatever you are flying, you're typically out in a region of the drag polar where a speed change will have little effect on the L/D ratio (which determines the descent angle) but will significantly change the descent rate.

Thus, I do believe using a constant angle assumption is a better approximation. You'll soon learn if you need to use 2.5 or 3.3 rather than 3, and the arithmetics can still easily be done mentally.

Cheers,
Fred

I thought I was doing good trying to avoid those airport hotels... and look at me now.

SlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 66
Reply 11, posted (11 years 12 months 2 days 21 hours ago) and read 9962 times:

I have always regarded mathematics as a wonderfully practical field. For that reason I am puzzled by this exercise.

A formula such as this might be very useful but yours starts with some absolutely preposterous conditions and utterly ignores the real world in which the question is posed.

For example, why on earth would any pilot ever want to maintain a constant horizontal speed? The 300 mph you give is simply not workable. At cruise it would be a mere Mach 0.38 (standard atmosphere) or roughly half the speed of prevailing traffic, and below 10000 MSL it would be an airspeed violation (and again entering the airport traffic area.) At the bottom of the descent it would be well above landing gear speed. Furthermore, horizontal speed is groundspeed and that is dependant on wind factor at each point during the descent, in other words, ignored by your formula.

OBTW, the descent point from 35000' for most jetliners is probably 90 to 105 nautical miles out depending on type and gross weight at TOD, which anyone with fifty hours in type could tell you from experience. That number is close enough because we get to play with the energy management most of the way down - turn points, adding drag and so on.

Happiness is not seeing another trite Ste. Maarten photo all week long.

Spencer From United Kingdom, joined Apr 2004, 1642 posts, RR: 15
Reply 12, posted (11 years 12 months 2 days 15 hours ago) and read 9902 times:

A rough guestimate is if you're at FL330, and you want to know your TOD point, in nm, then simply take the 33 from FL330, multiply by 3, and you have your 99nm TOD point. Works out pretty good on paper!
Spencer.

EOS1D4, 7D, 30D, 100-400/4.5-5.6 L IS USM, 70-200/2.8 L IS2 USM, 17-40 f4 L USM, 24-105 f4 L IS USM, 85 f1.8 USM

MITaero From United States of America, joined Jul 2003, 497 posts, RR: 7
Reply 13, posted (11 years 12 months 2 days 14 hours ago) and read 9887 times:

Guys -

it's a calculus exercise.. estimates and rules of thumb and other bits of info that aren't known to high-school teachers probably don't matter..

SlamClick From United States of America, joined Nov 2003, 10062 posts, RR: 66
Reply 14, posted (11 years 12 months 2 days 13 hours ago) and read 9880 times:

MITaero
Because all you can do with this exercise is generate a gee-whiz number that is completely unrelated to reality, I'd say it is the calculus exercise that does not relate to tech/ops.

If you are going to propose a mathematical question, why bother to pretend that it relates to the real world when the conditions themselves prevent that?

Happiness is not seeing another trite Ste. Maarten photo all week long.

MITaero From United States of America, joined Jul 2003, 497 posts, RR: 7
Reply 16, posted (11 years 12 months 2 days 7 hours ago) and read 9827 times:

It's an exercise meant to teach calculus, dude, that's all. Math books are full of these, but no one learns calculus until they complete them. It actually does help to put the problems in practical context just so students can understand the physical meanings of derivatives, integrals, etc.

haha, I had some extra paper left over from last summer (internship). i'm running out.

Yikes! From Canada, joined Oct 2001, 284 posts, RR: 1
Reply 17, posted (11 years 11 months 3 weeks 4 days 10 hours ago) and read 9696 times:

Neat discussion with regards "theory vs practise".

Me: (3 x altitude) + 10 nm for slowdown at 10K. Works like a charm, everytime.

BUT:Local knowledge comes to the forefront. Are they going to ask you to descend early? Are they going to ask you to slow down early? Are they going to ask you to hold?

In the execution of the low-drag approach, which commences at ToD, all these things must be taken into consideration.

Which brings one to the ONLY suitable philosophy in approach execution: LAR.