I would also like to program my graphing calculator (TI-82) to do the same
calculations. Can anyone suggest a source for the formulas or algorithms
to do these calculations?
Thanks in advance.
I've never heard of any formulas, but I think you've got a good idea. I've
started trying to make my own formulas. The only tricky ones involve wind,
TAS, etc. Speed and times ones are very simple : distance = speed * time.
So far, I've only derived a formula for wind correction angle. Given TAS, and
wind speed, your correction angle is : tan^-1 (wind speed/TAS). Note that
tan^-1 is inverse tangent, which is NOT really an inverse function. So, use
the inverse tan button - 2nd TAN.
I'll post more formulas as I get them.
International Baccalaureate Maxim of the Day
"Fifty lashes with a wet noodle!"
The above equation will only be correct for wind perpendicular to your
flight path True Course. It will increase in error as wind becomes a head or
tail wind (both of which should require 0 wind correction angle). The
formula becomes a little involved to put up here, but involves solving the
more general Pythogorean formula for triangles not limited to right angles
(i.e. C = SQRT(A^2 + B^2 - 2*A*B*COS(c) ) where
A, B, C are legs of the triangle,
c is the angle opposite leg C )
I've worked up a lot of these type equations in my line of work (flight
simulation). If anyone wants a more detailed explanation, feel free to
ask and I'll try and provide what help I can.
>I know this may not be much help, but it seems I once saw a book that
>was available that had formulas that could be used to program
>various computers to compute common E6-B flight computer functions.
>This book may be available from Sporty's or one of the aviation book
>clubs. This book was written with computers such as your TI-82 in
>mind. If I can find any additional info on this book, I'll pass it
Thanks. Such a book is just what I am looking for. Several people have
emailed me various formulas, but quite often they provide different
calculations for the same problem (ex. True Heading and Ground Speed). I
suspect that many of these are different approximations. The ones I could
duplicate on the TI-82 quite often give significantly answers than the
>Hey, E6B wanna-be's, let's start the net collection: here's my first
>contribution, the computation of a proper wind correction angle (WCA)
>given a desired course (TC) and the wind's direction and speed.
> WCA = ASIN( (WS/TAS)*SIN( WD-TC ))
>Where WS = Wind speed, and WD = Wind Direction, TC = True Course desired.
>You can then derive the groundspeed either from the resulting vector
>problem (knowing TH and TAS along with WD and WS) or using the law of
>cosines, which yields:
> GS = SQRT( TAS^2 + WS^2 - 2*TAS*WS*COS( TH-WD ) )
I haven't tried either of these specific algorithms, but I wrote a
flight planner that will do all of this stuff, and I used algorithms
based on the law of sines and the law of cosines. The algorithm I
use, however, takes in to account the quadrant each of the angles is
in, and alot of other +/- stuff. It's not pretty, but it gives
answers that are better than the ones I can obtain on the E6B,
considering the fact that the E6B has a reliability of less than 2
degrees (from the angle it is held, etc). Chances are there would be
difficulties if you were trying to fly nonstop across the country, but
I figure anybody who is willing to fly that far without using any
navaids or ground references deserves everything they get.
Anyway, my point is that if you use the above equations, don't get
discouraged if you get the wrong answer at first. Chances are, you
are just pointing towards the wrong quadrant.
Anybody who wants an even better algorithm should check out the
Great Circle Navigation Theory, which can be found in some book or
another. Chances are, when I get the energy and the desire for that
extra .5 degree or so accuracy, I will convert my algorithms over.
P.S. When I first was learning to fly, I tried to do the same thing
with my TI-85. I gave up, but it was something to do during History
Check out my new Web page at:
Put it on your hotlist, in fact. Visit often. Show it to
your friends. Wish it were yours.
Anyway, I am intensely interested in obtaining material that will
give me the relationship between atmospheric pressure, temperature
and altitude. I can calculate relatively easily, the rate and CG
problems but I need something a little more exact than graphs to
show me how to calculate density and pressure altitude. Enough
already with the generalisations (*aprox.* 2 deg. C drop for
*approx.* 1000ft of altitude gained, etc.)
I need to get formulas that will give me good values.
Thanks in advance!
I plotted the tabular data for pressure altitude correction and fit a line to
it figuring it would be easier to punch it in than look it up. The line was
basically linear but it didn't have a perfect fit. I didn't get real close
untill I was using a second order polynomial which became more of a chore to
punch in than to look up in the table. Oh well.
I reverse eng'd the table and ended up with
(Temps are in Deg C)
(Temp-(59-((pressure alt /1000)*3.34)))*65.6)+pressure alt
it gives me about the right values when compared to those elctronic
If anyone has the official one please post
In a world full of people only some want to fly,isn't that crazy
Gus Fraser, 212 235 0524. http://gti.net/fraser/flying.html
As an engineer, I know what it is to want to get precise answers, but ya
gotta watch it and not get carried away. Remember the old GIGO caveat!
You seem to 'slam' the E6B for not having greater than 2 degrees accuracy.
IF you are interested in theory, then by all means chase down that 0.5 or even
0.01 degree accuracy if you want (crank up the old double precision numbers).
Just don't forget, you are dealing with winds that are based on old measured data
and forecasts (we all know how accurate forecasts are!). There are no constant
winds in nature.
Also, trying to maintain any heading to within your goal of 0.5 should prove
rather interesting. I am reminded of the story of a military pilot undergoing
a checkride. The instructor pilot told the pilot to change heading 1 degree to
the left. The pilot answered back that 1 degree was ridiculous, how could he
be expected to change heading only 1 degree? The IP amended his instruction and
told the pilot to turn left 5 degrees. The pilot smugly replied that that WAS
doable and promplty changed heading. The IP then said, OK now turn right 4 degrees!
Keep up the good work on the equations, Shaun, but as for expecting any increase
in pilotage skills, I would guess that two pilots given the exact same route to
fly on the same day would not notice any difference regardless of whether a Cray
computer or an old E6B were used, IMHO :-)
2 degrees? So? Can you consistantly hold heading to +/- 2 degrees? If
you can, you're doing pretty good. In many cases, your instruments may
not even be that good. Besides, if the winds aloft forcasts are any
better than +/- 10 degrees and +/- 20% in speed, that's unusual too.
Roy Smith <r...@nyu.edu>
Hippocrates Project, Department of Microbiology, Coles 202
NYU School of Medicine, 550 First Avenue, New York, NY 10016
"This never happened to Bart Simpson."
The temps are of course in Deg F.
Determine your pressure altitude.
Determine standard temperature for your pressure altitude.
Determine the difference: actual temp - standard temp.
Multiply this difference by 125 (or figure 1000ft/8C).
Add this to your pressure altitude. This is your
Assume your pressure altutide is 7500 and temperature is 18C.
Pressure alt = 7500.
Std tmp = 15 - (2)(7.5) = 15 - 15 = 0
difference = 18 - 0 = 18
18 x 125 = 2250
7500 + 2250 = 9750 = density altitude.
> Anyway, I am intensely interested in obtaining material that will
> give me the relationship between atmospheric pressure, temperature
> and altitude.
Here's a physicist's version, please ask if I use units/terms you aren't
It uses K as unit of temperature (0 degC = 273.15 K, 100 degC = 373.15 K).
Using Z to represent vertical distance (usually altitude, thus Z = 0
normally represents sea level).
For a dry atmosphere with constant lapse rate (i.e. T = T0 - a*Z) where a
is the lapse rate (rate of temperature fall per unit height) and T0 is the
(thermodynamic/absolute) temperature at Z = 0:
P/P0 = (T/T0)^(b/a) ^ is to-the-power-of
where P0 is the pressure at Z = 0 (the altimeter setting if Z = 0) and b
is a constant. Note that this is the T defined above, so the expression
is really P as a function of Z, with T as a convenient parameter including
b = molar-mass-of-air * g / R which works out (in convenient units) to
about 10.5 K per 1000 ft if I remeber correctly, but check it.
This works for any atmosphere in which temeperature varies linearly with
height (except a = 0 of course), but the following give the conventional
values of a.
The International Standard Atmosphere uses a = 1.98 K per 1000 ft (6.5 K
per 1000 m). T0 = 288.15 K. I believe altimeters are calibrated to this.
The Jet Standard Atmosphere uses a = 2 K per 1000 ft.
Note that *real* atmospheres are usually nothing like this, with very
wiggly T vs Z graphs.
>2 degrees? So? Can you consistantly hold heading to +/- 2 degrees? If
>you can, you're doing pretty good. In many cases, your instruments may
>not even be that good. Besides, if the winds aloft forcasts are any
>better than +/- 10 degrees and +/- 20% in speed, that's unusual too.
Oddly enough, that was exactly my point. I didn't mean to suggest
that an E6B was unreliable. Quite the opposite, really. What I was
saying was that the errors in non-accounting for the curvature of the
Earth are less than the errors that you would get if you were holding
the E6B at the wrong angle, all of which are sufficiently negligable.
Sorry about the misunderstanding,
I am quite familiar with the S.I. system of Units and much more so, in
fact. I went to school in Nigeria, West Africa and that's all we used
in Physics and Chemistry. I liked Physics so much that I am now in my
last year as an Electrical Engineering student! :)
Anyway, I did try to work out the relationship b/w alt., temp., and
pressure. I used the General Gas Equation and Charle's Law but gave it
up when I discovered it'd take longer than I thought. Laziness!
However, I strongly suspect that if I'd followed it to it's logical
conclusion, I'd have gotten something worked out.
Here then, just for the heck of it is what I started out with:
P1=Atmospheric Pressure at sea-level
P2=Atmospheric Pressure at height, h.
T1=Temp. at sea-level
T2=Temp. at height, h.
Question: Will this work if P1=height at airport? and P2, altitude
height? I should think so.
Pressure's relationship with height:
rho=density of air
g=acceleration due to gravity.
Now, I am sure that with a little algebra, I could figure out a way to
calculate pressure altitude If I knew all those values above. Okay,
now, I know the pressure altitude; how to get from there to density
altitude? I know that density altitude is pressure altitude corrected
for temperature effects. What is the relationship?
Also, where did you get the equation, p/p0=(t/t0)^(b/a)? Would like to
know. What to do when lapse rate is not constant? Or am I going to
have to live with approximations? Damn you, mama nature!
P(z) = Po*e^(-mgz/kT)
T=Temp in Kelvins
P(z)=pressure at z meters
Po=sea level Pressure
m=mass of one molecule of air (approx 29 grams per mole, convert for yourself)
g=acceleration of gravity = 9.80665 m/s/s
k=Boltzmann's constant = 1.380658*10^-23 J/K
\ \ No bird soars too high,
\ \ If he soars with his own wings.
|\ \ \ - William Blake
| \----\ \------\
| ----- ------/
| / / /
|/ / /
Unfortunately, that does not work because it assumes that the *density* is
the same at both levels, which it is not.
> Pressure's relationship with height:
> rho=density of air
> g=acceleration due to gravity.
Again, you're assuming constant density in using this equation. For every
infinitesimally thin layer of atmosphere it is appropriate, so what you
end up with is a *differential* equation (the hydrostatic equation)
dP/dh = rho*g (1)
where rho is a function of h
> Also, where did you get the equation, p/p0=(t/t0)^(b/a)?
I integrated the hydrostatic equation. First you setup your lapse rate
T = T0 - a*h (2)
Then you use the ideal gas law to say
rho = Mm * P/(R*T) (3)
where Mm is the molar mass (about 0.029 kg mol^-1 for air) and R is the
gas constant (8.314 J K^-1 mol^-1).
Putting (2) and (3) in (1) gives you
dP/dh = (Mm * g / R) * P/(T0 - a*h) (4)
= b * P/(T0 - a*h) in my terminology, where b is a constant.
[Incidentally, if a = 0 and you integrate you get the equation that Arun
gives in article <aprakash-270...@pm233.cloudnet.com>, but
generally that's not the case in the atmosphere.]
Integrate it (with P = P0 at h = 0) and you get
P/P0 = (1 - a * h / T0)^(b/a) (5)
P/P0 = (T/T0)^(b/a) (6)
using (2) again for convenience.
Note that if you do use 6, T is just a convenience parameter, and the
expression is still really P as a function of h. For (6) to be exact it
requires the temperature variation to be linear up with h between T0 and
> What to do when lapse rate is not constant? Or am I going to
> have to live with approximations? Damn you, mama nature!
Either get better at integration than me, or do it numerically! You can
build up successive layers with different but constant lapse rates if you
> now, I know the pressure altitude; how to get from there to density
> altitude? I know that density altitude is pressure altitude corrected
> for temperature effects. What is the relationship?
Well using (3) in (6) you get:
rho/rho0 = (T/T0)^(b/a - 1) (7)
That gives the density rho as a function of h. An exact expression for
the density *altitude* is, as far as I can see, horrible, but more often
you're not interested in the density altitude as such, just the density.
For example, for airspeeds:
TAS/CAS = (rho0/rho)^(1/2) = (T/T0)^((1 - b/a)/2) (8)
Hope that helps.
Adding all errors of weather forecasting, instruments, E6B errors, etc.
you'll end up in the wrong country without navaids or a sectional.
International Baccalaureate Maxim of the Day
"Eye warship satin."
I learned to fly in France, and was taught to do wind correction and
time/distance problems in my head, using a system of little tricks (you
start with what's called the "base factor" = 60/TAS and go on from there).
You have to be pretty good at mental arithmetic (few Americans are; the
French were mostly pretty good) but if you are it works really well and you
don't need any E6B or calculator. The answers you get can be off by a degree
or two, but as several people have pointed out, that's minor compared to the
other possible errors (especially in winds aloft). Have any of the Europeans
(I'll include the British here) on this newsgroup learned a similar system?
When I learned in the mid 1980s, all French student pilots were taught this
method as far as I could tell.
If people are interested, I'll post some notes I've written up.
[ snip ]
>If people are interested, I'll post some notes I've written up.
Barry.. I'm interested in anything that reduces cockpit workload.
Please post the notes or e-mail to me.
Thanks in advance,
>If people are interested, I'll post some notes I've written up.
Barry... I'm interested in anything that helps reduce cockpit workload.
Please post the notes or e-mail them.
Thanks in advance,
While it may be true that Americans are not good at 'Rithmetic, we
don't like to be told in such brash terms about it! The French
good at a *lot* of things including some I know who have a couple
things to know about bathing and rudimentary Hygenic Theory!
<There, I told him>
That being said, I would love to know what your "French" methods are.
Frederic le trois!
By popular demand, here are my notes on the method I learned in France. And
no, I am not French.
Barry H. Silverman
Mental calculations for cross-country flight planning
Student pilots in France learn to calculate true course and time
enroute mentally, without an E6B or electronic calculator. This
method requires proficiency in mental arithmetic but works
amazingly well with practice. The errors introduced are
negligible when one considers the uncertainties in winds aloft
The first step is to calculate what we will call the base factor,
designated F. To get F, just divide 60 by the true airspeed in
knots. For a particular airplane, F is almost constant, though
it will vary somewhat with the choice of cruise power and density
altitude. For a Piper Tomahawk, typical cruise speed is about 90
kt, so F = 60/90 = 2/3. For a Skyhawk or Archer that cruises at
120 kt, F = 1/2.
The first use of F is to calculate estimated time enroute (ETE)
with no wind. Simply take the distance in nautical miles and
multiply by F to get the time enroute in minutes.
Example: To travel 36 nm in a Tomahawk at 90 kt will take
(2/3) x 36 = 24 min if there is no wind.
To correct for the wind, we first multiply the speed (in knots)
by F. We will call this number the wind factor W.
Example: If the winds aloft are from 320 degrees at 20 knots,
in a Tomahawk the wind factor W = (2/3) x 20 ÷ 13.
The wind factor W is used in two different ways. W will be the
maximum wind correction angle for a direct crosswind. In the
example above, the wind direction is 320. If our desired true
course (TC) is 230 (perpendicular to the wind), then we will have
to correct into the wind by 13 degrees (all of W), giving us a
true heading TH = 230 + 13 = 243. If the wind is not a direct
crosswind, we take the crosswind component of W. In mathematical
terms, we must multiply W by sin(A), where A is the angle between
the wind and the desired course. As a practical matter, we will
just estimate sin(A) by remembering a few key values. If the
angle A is 20 deg, multiply W by 1/3. For 30 deg, take W x 1/2.
For 50 deg, take W x (3/4), and for anything above 60 deg, take
all of W as the crosswind component. If the angle A is greater
than 90 deg (a tailwind instead of a headwind), replace A by
(180 - A) and then use the same approximations.
Example: If, as above, wind is from 320 deg at 20 kt, and our
desired course TC = 092, we have a left crosswind with the
angle A = 132, so we use A = 180 - 132 = 48 instead. This is
almost 50 deg, so we use a factor of 3/4 and take W x (3/4) =
13 x (3/4) ÷ 10 deg as the crosswind correction and find TH =
092 - 010 = 082.
The second use of the wind factor W is to correct the no-wind
time enroute already calculated. We take the head or tailwind
component of W and call this the time correction factor T. This
becomes the number of minutes we must add or subtract for each
hour enroute. The headwind component will be
W cos(A) = W sin(90 - A), so we can use the same approximations
as above, but with 90 - A in place of A.
Example: As above, wind is 320 at 20, TC = 092, F = 2/3, so W
= 13. The angle A = 48, so (90 - A) = 90 - 48 = 32, or close
to 30, and we take 1/2 of W as our time correction factor T.
So for each hour of flight, we must subtract (because we have
a tailwind) 1/2 x W = 1/2 x 13 ÷ 6 minutes. If, as before,
the distance is 36 nm, the no-wind time is F x 36 = 2/3 x 36 =
24 min, or a little less than half of an hour. So we subtract
half of 6 minutes, or 3 minutes, from the no-wind time of 24
min, to get ETE = 21 min, to go with our true heading
TH = 082. Doing the problem exactly on an electronic
calculator gives TH = 082.5 and ETE = 21.1 min.
There is a complication when the time factor T is greater than
about 6. The method will tend to overestimate a tailwind
correction, and underestimate a headwind correction, and we
should make a small additional correction for this. A correction
table exists, but it is usually sufficient to just add an extra
minute or two (for a headwind), or subtract a minute or two (for
a tailwind), if T is large. The method also makes no time
correction for a strong crosswind, which reduces groundspeed
slightly, and so if the wind correction angle is very large (over
15 or so) you can adjust the time correction factor slightly.
This sounds imprecise, but remember that the winds aloft given in
a briefing are estimates. The uncertainties in winds aloft and
true airspeed will usually result in errors greater than those
introduced by our approximations.
Example: Suppose that in a Tomahawk (F = 2/3) we have a
direct crosswind of 30 kt. With our method, no time
correction would be made for this, and we would estimate the
time for a 36 nm leg as 24 min, when in fact the actual time
would be 25.5 min. The error is 1.5 min, which is still
fairly small. But now consider what would happen if the winds
aloft forecast was off by only 10 degrees, so that we now have
a quartering tailwind instead of a direct crosswind. The
actual time would now be 23.9 min, so the error introduced by
the winds being off by only 10 degrees is greater than the
error of our calculation! So while the method might
occasionally be not quite precise enough for an FAA written
exam, it is fine for real flying.
We can make a couple of observations here:
1) This method bypasses completely the step of calculating the
groundspeed. This is fine, because we never really care about
the groundspeed itself, but only use it to calculate time
2) By separating the time calculations into a no-wind time plus a
correction, more of the flight planning can be done early. The
no-wind times can be estimated and placed on the log without any
knowledge of the winds aloft. After the winds aloft have been
obtained, only the corrections need be made. This reduces the
time needed on the day of the flight for planning.
Let's do one more complete example to summarize the method. We
will take wind aloft as 280 at 24 kt, TAS = 120 kt, true course
TC = 247, and a leg length D = 37 nm.
1) Base factor F = 60/120 = 1/2.
2) No-wind time = F x D = 1/2 x 37 = 18.5 min.
3) Wind factor W = F x wind = 1/2 x 24 = 12.
4) We have a right quartering headwind, with the crosswind angle
A = 280 - 247 = 33 deg. For 33 deg we want just over half of W
as our wind correction angle, so 1/2 x 12 = 6 deg will be our
wind correction, and TH = 247 + 6 = 253.
5) For the headwind, we have an angle of 90 - 33 = 57 deg. For
50 deg we would take 3/4 of W = 3/4 x 12 = 9, so for 57 deg we
want slightly more as the time correction T. In addition,
because T is fairly large, we need to take a little more, so
we'll add 11 minutes per hour flown. The no-wind estimate was
18.5 min, which is a little less than a third of an hour, so
we'll add 1/3 x 11 or 3.5 min. So the ETE = 18.5 + 3.5 = 22 min.
So our mental method estimated TH = 253 and ETE = 22 min. A
precise calculation gives TH = 253.3 and ETE = 22.4 min.