Folks,
Tom's note copied below, seems to be somewhat relevant to this thread,
and because you'all are by all accounts in no way math phobic,
you may enjoy a minor wrestle with this physical formulation.
But above all, Tom seems to have developed a quite inflated
conception of my aero physics insights (though by all accounts,
his other mailee, Doctor John Lowry is much more worthy.
(He has received grants for developing usable methods for
evaluating light aircraft performance using his
physics insights, Ref#1)
If you can see through Tom's difficulty (as I am confident you will)
then I would appreciate a copy of your response to the list, or to me
if you think it otherwise inappropriate.
Thank you.
Brian
Ref: A sample of John Lowry's works: (Self-published):
Computing Airplane Performance - A Field Guide (1995)
J.T Lowry
Flight Physics
PO Box 20919
Billings
Montana 59101
tel: 406 - 248-2606
Date: Wed, 11 Aug 1999 09:43:44 -0500
From: Tom Turton <tturton@ntx.waymark.net>
Organization: Raytheon ITSS
X-Accept-Language: en
To: john lowry <jlowry@mcn.net>, brian whatcott <inet@intellisys.net>
Subject: Temperature correction on altitude, specifically Gamma
John, Brian -
You guys have been great to bounce aero equations off of, and I
seek you expertise again :-)
I'm wrestling with an equation someone else generated to compute
flight path angle (gamma) corrected for temperature effects.
Actually, the only reason they did this (AFAIK) was because they
are NOT using true, geometric altitudes, but rather pressure altitudes.
From all my reference sources, and working through it myself also,
I agree that (neglecting thrust angle of incidence) you get:
(and let's assume small angle approximation so I don't have to
keep writing sin/arcsin....just let sin_gamma be replaced by gamma)
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:
That "looks more better" to me, yet plugging numbers in it doesn't
give me a warm fuzzy. I'm not seeing what I feel I should, and
this is what I really want to run by you guys.
Picked some numbers out of the air:
T = 6000
D = 3861.4
W = 36000
(Vt/g)*(d_Vt/d_h) = 0.13507 (acceleration term)
Plugging these into eqn 1 gives gamma = 0.05234 radians = 3 degrees
Now, if we use one of our hot, Texas days: T_act = 100 deg F
or 559.7 deg R and T_std = 518.7 deg R, so d_hp/d_hg = 0.9267
Now, using eqns (2) and (3) (note: assuming that conditions were
such that the acceleration term using d_Vt/d_hp was still 0.13507)
then, gamma_1 = 3 degrees and gamma_true = 3 * 0.9267 = 2.78 degrees
This jumped out at me as being wrong from all our previous discussions
on temperature and altitude. Hot days equate to "expanded" air
columns. I.e. climbing to 10,000 feet indicated on our 100 degree
day really means you are at about 10,790 geometric. Therefore,
it seems to me that if eqn (2) told you that your gamma was 3 degrees
in climbing to 10,000 feet indicated, your true gamma would have
to be steeper because you were REALLY closer to 10,790 feet. Make
sense?
But when I use what I believe is the correct gamma equation (eqn 4)
I only get a gamma slightly larger (3.03 degrees). This doesn't
seem to jive because if I sketch out a triangle showing a 3 degree
gamma climb angle to 10,000 feet I would have the altitude as 10,000
(duh) and the ground path distance as 190811 feet. BUT, if I am
REALLY at 10,790 feet true (rather than my 10000 indicated) then the
math says my true gamma is more like 3.23 degrees.
So guys, what gives? Which calculation would YOU believe and why?
Also, secondary question regarding the temperature correction factor
itself (d_hp/d_hg = T_std/T_act).... Does it make a difference
WHERE the temps are measured? Typically, I've seen them given as the
temperatures at the ground (sea level or reporting station elevation),
but was wondering if you could use the temperature at altitude?
And I guess implicit in all of this is that your temperature lapse
rate is the standard one used and that you don't have any kind of
inversion, or much larger/smaller lapse rate than standard.
I'm going nuts here and hope one or both of you can pull me out
of this equation muck! Thanks in advance.