508 views

Skip to first unread message

Jan 10, 1996, 3:00:00โฏAM1/10/96

to

I just bought the Slide Graphic Flight Computer by Jeppesen. I never

liked slide rules but thought it might be useful to have such a calculator

that didn't depend on batteries.

liked slide rules but thought it might be useful to have such a calculator

that didn't depend on batteries.

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.

Jan 10, 1996, 3:00:00โฏAM1/10/96

to

Jan 14, 1996, 3:00:00โฏAM1/14/96

to

In article <4d1arc$e...@dartvax.dartmouth.edu>, tom.j...@valley.net says...

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.

--James Rankin

mo...@nrv.net

Salem, Virginia

International Baccalaureate Maxim of the Day

"Fifty lashes with a wet noodle!"

From SHS

Jan 15, 1996, 3:00:00โฏAM1/15/96

to

> 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).

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.

---Tom T.

Jan 17, 1996, 3:00:00โฏAM1/17/96

to

mo...@nrv.net (James Rankin) wrote:

>

> In article <4d1arc$e...@dartvax.dartmouth.edu>, tom.j...@valley.net says...

> >

> >I just bought the Slide Graphic Flight Computer by Jeppesen. I never

> >liked slide rules but thought it might be useful to have such a calculator

> >that didn't depend on batteries.

> >

> >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.

> >

>

>

> In article <4d1arc$e...@dartvax.dartmouth.edu>, tom.j...@valley.net says...

> >

> >I just bought the Slide Graphic Flight Computer by Jeppesen. I never

> >liked slide rules but thought it might be useful to have such a calculator

> >that didn't depend on batteries.

> >

> >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

> 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.

>

> --James Rankin

> mo...@nrv.net

> Salem, Virginia

> International Baccalaureate Maxim of the Day

> "Fifty lashes with a wet noodle!"

> From SHS

>

I know this may not be much help, but it seems I once saw a book 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.

>

> --James Rankin

> mo...@nrv.net

> Salem, Virginia

> International Baccalaureate Maxim of the Day

> "Fifty lashes with a wet noodle!"

> From SHS

>

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

along.

Jan 18, 1996, 3:00:00โฏAM1/18/96

to

Rex Cox <r...@icon.net> wrote:

>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

>along.

>

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

E6-b gives.

Jan 18, 1996, 3:00:00โฏAM1/18/96

to

Jan 18, 1996, 3:00:00โฏAM1/18/96

to

Jan 25, 1996, 3:00:00โฏAM1/25/96

to

p...@apple.com (Patrick W. P. Dirks) wrote:

snip, snip

>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.

snip snip...

> 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.

good luck,

Shaun

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

class.

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

Check out my new Web page at:

http://ccwf.cc.utexas.edu/~shaun

Put it on your hotlist, in fact. Visit often. Show it to

your friends. Wish it were yours.

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

Jan 25, 1996, 3:00:00โฏAM1/25/96

to

I remember asking the group some time ago questions pertaining to

this same topic but did not receive any answers. I wonder why?

this same topic but did not receive any answers. I wonder why?

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!

Jan 26, 1996, 3:00:00โฏAM1/26/96

to

In article <31080E...@oe.fau.edu>, fwoo...@oe.fau.edu says...

:

:I remember asking the group some time ago questions pertaining to

:

:I remember asking the group some time ago questions pertaining to

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.

Jan 26, 1996, 3:00:00โฏAM1/26/96

to

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

versions.

If anyone has the official one please post

---

---------------------------Seal----------------------------------

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

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

Jan 26, 1996, 3:00:00โฏAM1/26/96

to

In article 9...@geraldo.cc.utexas.edu, sh...@mail.utexas.edu (Shaun Dawson) writes:

> p...@apple.com (Patrick W. P. Dirks) wrote:

> >

> snip snip...

>

> > WCA = ASIN( (WS/TAS)*SIN( WD-TC ))

>

> >Where WS = Wind speed, and WD = Wind Direction, TC = True Course desired.

>

> p...@apple.com (Patrick W. P. Dirks) wrote:

> >

> snip snip...

>

> > WCA = ASIN( (WS/TAS)*SIN( WD-TC ))

>

> >Where WS = Wind speed, and WD = Wind Direction, TC = True Course desired.

>

> > 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.

[snip]>

> 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.

> 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.

>

Shaun,

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 :-)

---Tom T.

Jan 26, 1996, 3:00:00โฏAM1/26/96

to

sh...@mail.utexas.edu (Shaun Dawson) wrote:

> 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

> 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

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."

Jan 26, 1996, 3:00:00โฏAM1/26/96

to

Sorry,

The temps are of course in Deg F.

woops.

Jan 27, 1996, 3:00:00โฏAM1/27/96

to

Here is a GOOD approximation that most people can do in their head.

First, remember that at sea level, standard conditions require 15C and

the standard lapse rate is 2C/1000ft. So:

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

density altitude.

Example:

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.

--

Boring signature follows:

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

Gary J. Blair PP-ASEL (510) 294-3819 workdays

gjb...@netcom.com (209) 836-5773 other occasions

Jan 27, 1996, 3:00:00โฏAM1/27/96

to

In article <31080E...@oe.fau.edu>, fwoo...@oe.fau.edu wrote:

> 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

familiar with.

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

Z.

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.

Julian Scarfe

ja...@cus.cam.ac.uk

Jan 27, 1996, 3:00:00โฏAM1/27/96

to

r...@popmail.med.nyu.edu (Roy Smith) wrote:

>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,

Shaun

Jan 27, 1996, 3:00:00โฏAM1/27/96

to Julian Scarfe

Thanks Julian S.

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/T1=P2/T2=n

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:

h*rho*g=P

h=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!

Thanks!

Yours,

Frederic

Jan 27, 1996, 3:00:00โฏAM1/27/96

to

The pressure P at height Z at temp T is defined below:

P(z) = Po*e^(-mgz/kT)

T=Temp in Kelvins

P(z)=pressure at z meters

Po=sea level Pressure

e=2.718281828

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

--

Arun Prakash

apra...@cloudnet.com

__

\ \ No bird soars too high,

\ \ If he soars with his own wings.

|\ \ \ - William Blake

| \----\ \------\

| ----- ------/

| / / /

|/ / /

/ /

/_/

Jan 28, 1996, 3:00:00โฏAM1/28/96

to

In article <310AA1...@oe.fau.edu>, fwoo...@oe.fau.edu wrote:

> 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/T1=P2/T2=n

>

> 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.

> 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/T1=P2/T2=n

>

> 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.

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:

>

> h*rho*g=P

>

> h=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

relationship:

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)

or

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

T.

> 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

prefer.

> 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?

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.

Julian Scarfe

ja...@cus.cam.ac.uk

Jan 28, 1996, 3:00:00โฏAM1/28/96

to

I asked this question about flight computer formulas a while back. The

best thing I have found so far is a file in the AVSIG forum on CompuServe.

It is about 5 pages. I will post it here after getting permission from

the original author.

best thing I have found so far is a file in the AVSIG forum on CompuServe.

It is about 5 pages. I will post it here after getting permission from

the original author.

Tom Jacobs

Jan 28, 1996, 3:00:00โฏAM1/28/96

to

In article <roy-260196...@mchip8.med.nyu.edu>, r...@popmail.med.nyu.edu

says...

>

>sh...@mail.utexas.edu (Shaun Dawson) wrote:

>> 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

>

says...

>

>sh...@mail.utexas.edu (Shaun Dawson) wrote:

>> 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

>

>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.

>

>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."

>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."

--

Adding all errors of weather forecasting, instruments, E6B errors, etc.

you'll end up in the wrong country without navaids or a sectional.

--James Rankin

mo...@nrv.net

Salem, Virginia

International Baccalaureate Maxim of the Day

"Eye warship satin."

From SHS

Jan 29, 1996, 3:00:00โฏAM1/29/96

to

In article a...@kirk.nrv.net, mo...@nrv.net (James Rankin) wrote:>

> --

> Adding all errors of weather forecasting, instruments, E6B errors, etc.

> you'll end up in the wrong country without navaids or a sectional.

>

> --

> Adding all errors of weather forecasting, instruments, E6B errors, etc.

> you'll end up in the wrong country without navaids or a sectional.

>

Boy,

It's a good thing Charles Lindbergh didn't know that, huh?!

Jan 29, 1996, 3:00:00โฏAM1/29/96

to

On the subject of formulas for wind correction, etc.:

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.

Barry

Jan 30, 1996, 3:00:00โฏAM1/30/96

to

In article <4eijp9$h...@hacgate2.hac.com>, ttu...@samson.hac.com says...

--

Some of the time, he got valuable outside references. I was refering to IMC.

--James Rankin

mo...@nrv.net

Salem, Virginia

International Baccalaureate Maxim of the Day

"I am not evil. I am not the Devil."

From SHS

Jan 31, 1996, 3:00:00โฏAM1/31/96

to bsilv...@admin.tc.faa.gov

bsilv...@admin.tc.faa.gov (Barry H. Silverman) wrote:

>On the subject of formulas for wind correction, etc.:

>

>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).

>On the subject of formulas for wind correction, etc.:

>

>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).

[ snip ]

>

>If people are interested, I'll post some notes I've written up.

>

>Barry

>

Barry.. I'm interested in anything that reduces cockpit workload.

Please post the notes or e-mail to me.

Thanks in advance,

John

Jan 31, 1996, 3:00:00โฏAM1/31/96

to

bsilv...@admin.tc.faa.gov (Barry H. Silverman) wrote:

>On the subject of formulas for wind correction, etc.:

>

>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).

>On the subject of formulas for wind correction, etc.:

>

>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).

[snip]

>If people are interested, I'll post some notes I've written up.

>

>Barry

>

Barry... I'm interested in anything that helps reduce cockpit workload.

Please post the notes or e-mail them.

Thanks in advance,

John

Jan 31, 1996, 3:00:00โฏAM1/31/96

to

Gee, Barry Silverman:

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.

Thanks,

Frederic le trois!

Feb 2, 1996, 3:00:00โฏAM2/2/96

to

In article <310FA3...@oe.fau.edu>, fwoo...@oe.fau.edu says...

>

>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.

>

>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.

By popular demand, here are my notes on the method I learned in France. And

no, I am not French.

===========================================================================

Barry H. Silverman

October 1994

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

forecasts.

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 รท 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) รท 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 รท 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

enroute.

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.

Reply all

Reply to author

Forward

0 new messages