Chronology Current Month Current Thread Current Date
[Year List] [Month List (current year)] [Date Index] [Thread Index] [Thread Prev] [Thread Next] [Date Prev] [Date Next]

Re: Misconceptions: Physics of Flight

Folks,
John Lowry copied me with his response - far too quickly for my taste!
:-)

I summarize Tom's original question beneath.

Sincerely
Brian
------------------------------------------------------------------------
From: "John T. Lowry" <jlowry@mcn.net>
To: <PHYS-L@lists.nau.edu>
Cc: <inet@intellisys.net>, <tturton@ntx.waymark.net>
Subject: Temperature correction for gamma
Date: Wed, 11 Aug 1999 16:34:27 -0600


August 11, 1999

Hi Tom, Brian, and ALL:

Tom and I go back quite a ways as regards wrestling with non-standard
atmospheres. They always give me a headache. I think that is for the
standard "complicated problem" reason: too many ways to look at it...which
one (or more!) is correct?? Still, I'll try to parse more or less step by
step, then make some comments. The small flight path angle approximation
will be assumed, as Tom laid down, throughout.

Eqn. 1 is correct and, unless it's a standard day, eqn. 2 is not correct.
The question is: How to correct eqn. 2 so it's almost correct. There's no
need to, since eqn. 4 is correct. (Simply by substitution of the correct if
unnumbered equation cited by Tom between his eqn. 2 and his eqn. 3.) I think
Tom meant to write 'denominator' instead of 'numerator.'

But now comes the problem with "sensible" interpretation. For starters, jump
to the end of Tom's note, the question about WHERE to measure T_std/T_act.
Answer: LOCALLY! That is, at the airplane's location. That's not much
problem if you assume you know the pressure and temperature locally. (OK,
you may have to crank in a ram-correction effect for the temperature; but
you can always get the local pressure just by reading the altimeter -- if
you've got my new -- AD! AD! -- American Institute of Aeronautics and
Astronautics book Performance of Light Aircraft, ISBN 1-56347-330-5, 515
pp., $64.95, and look in the first two chapters.) By the way it was Tom who
gave me the idea of showing that pressure altitude differences are the same
as indicated altitude differences. Thanks, Tom!

So if you're monitoring d_Vt/d_hp you can correct with the temperature ratio
T_std/T_act (less than unity if today's atmosphere is everywhere hotter than
standard) and get the veridical eqn. 1. And Tom's correct that an indicated
10,000 ft would be (same lapse rate, same standard pressure) a tape line
10,790 ft. But there's nothing that says that the airplane would climb
faster under these conditions! The above-cited AIAA book has a section, near
the end of Chapter 1, discussing all this in a time-to-climb context. The
example is too long to give here, but the selection below may give the
flavor of it:

"Does that mean, translated to the standard atmosphere, that this climb
would have been from density altitude 2318' up to density altitude 3394', a
total of 1076', in actual elapsed time 50 seconds? Yes the 1076', but not
the 50 seconds!

"Here's why not. Rate of climb depends on air density, but the obstacle
being overcome is not density. If it were, if the airplane were climbing up
density steps or rungs, one molecule at a time (fixed time for each step),
then 50 seconds for the trip would make sense. But that's not the problem.
It's the gravitational force which has to be overcome and the actual brute
tapeline (geopotential) distance really matters."

I go on to calculate rate of climb and show the needed time interval is
actually about 56.5 sec.

Also in the AIAA book is a detailed discussion of "non-standard linear
atmospheres," i.e., those with constant lapse rates but with non-standard
VALUE of lapse rate and/or of MSL pressure and/or of MSL temperature. Most
discussions on this kind of problem fall apart into cacophony because a
through-going model of "today's" atmosphere is not agreed upon, beforehand,
by the participants. My discussion shows how one goes about deriving all of
today's necessary atmospheric constants from temperature and pressure
information at the airplane and at the weather station (assumed directly
beneath the airplane at a given field altitude). It's perhaps a little
messy, but it does have the virtue of always giving correct answers (to
well-posed problems).

One last sniping comment. The equation Tom (very reasonably) questioned,
eqn. 3, is a horrible example of analytical thinking. Or, if one prefers to
be "positive," a wonderful example of fuzzy thinking. Just because some
variable in a formula varies according to some given factor CERTAINLY
doesn't mean that the WHOLE FORMULA varies according to that same factor.
(Unless fortuitously.) Or what's a calculus for?? You can't throw finagle
factors around ad libitum!

Anyway, it was good to hear from you guys. I hope this helps.

John.

John T. Lowry, PhD
Flight Physics
Box 20919; Billings MT 59104
Voice: (406) 248-2606
jlowry@mcn.net



------------------------------------------------------------------


At 12:14 8/11/99 -0500, Tom Turton wrote off-list:

I'm wrestling with an equation someone else generated to compute
flight path angle (gamma) corrected for temperature effects.

...just let sin_gamma be replaced by gamma)

(T-D)
gamma = ------------------------------ (eqn 1)
W [ 1 + (d_Vt/d_hg)*(Vt/g) ]

T = Thrust
D = Drag
W = Weight
g = gravitational acceleration
Vt = true velocity
hg = geometric altitude
d_ = derivative

Now, if one has instead been using PRESSURE altitude (actually,
altimeter altitude) you would have:

(T-D)
gamma_1 = ------------------------------ (eqn 2)
W [ 1 + (d_Vt/d_hp)*(Vt/g) ]


I'm sure you all have seen (or even posted to newsgroup) the
standard correction formula:

d_hp/d_hg = T_std/T_act

where: T_std = standard atmosphere (absolute) temperature
T_act = actual (absolute) temperature
d_hp = change in pressure altitude
d_hg = change in geometric altitude

Now the equation I'm questioning does this:

gamma(true) = gamma_1 * (d_hp/dhg) (eqn 3)

I felt it would be more accurate to apply the correction to the term
in the numerator:

(T-D)
gamma = ----------------------------------------- (eqn 4)
W [ 1 + (d_Vt/d_hp)*(d_hp/d_hg)*(Vt/g) ]


That "looks more better" to me, yet plugging numbers in it doesn't
give me a warm fuzzy. ...
---Tom Turton


brian whatcott <inet@intellisys.net>
Altus OK