Gliding flight - maximum range & wind effect 2

[khlama]

Could somebody help me with this one please:

How do you account of a head or tail wind while maximizing the range of a glider?

I know that in order to maximize the range of a glider we should fly at the minimum drag speed. But the minimum drag speed is usually computed for a still air. And thus the range that we usually maximize is an "air range" not a "ground range".

How should the true air speed be selected in order to maximize the "ground range" that will allow us to account of any wind effect.

Adding or substructing the wind speed to the "still-air" minimum drag speed seems a bit too obvious to me...
A relation may exist between the wind speed, true air-speed and minimum speed for still-air flight... but what would be the steps to obtain it.

Many thanks in advance for your advices

regards

 

[berkshire]

The normal method of maximizing range with a glider is to fly at minimum sink speed when flying down wind, allowing the wind to carry you over the ground while maximizing your time aloft.
Going into the wind is a little more difficult, here you fly at your best L/D speed then add the speed of the wind over the ground to your airspeed. Although this increases your rate of sink you will cover more ground.
It should be noted that the critical speeds are not minimum drag; they are the spots on the polar curve where the wing generates the highest coefficient of lift, min sink.
And max lift to drag ratio which may be well less than max lift.

 

[rb1957]

um, er, maybe it isn't obvious but ...
airspeeds are with respect to still air,
and ground speeds are with respect to the ground;
so the equivalent ground speed is the vector sum of the airspeed and the wind speed (airplanes land and take-off into the wind to minimise their ground speed.

 

[khlama]

thank you berkshire and rb1957.
berkshire your posting makes sense. But while flying at the best L/D and for small descent path angles we can assume that the Lift will be equal to the glider weight. Hence the best L/D will occure at the minimum Drag. We can take then the derivate of the Drag (assuming a parabolic polar) with respect to the speed to vanish at the minimum drag speed. But again this speed will represent the maximum "air" range.
To maximize the ground range... we cannot only consider the ground speed instead of the air speed, because this will only lead to Vmd + Vw (minimum drag speed + wind speed) wish does not necessary maximize the ground range.

I know something must be considered as well in order for all to work out... but I'm clueless, and there isn't that much information on that topic on the net...

Many thanks in advance for your help

 

[berkshire]

Gentlemen,
As you point out my answer did not make complete sense.
As a glider pilot I was confusing technique with theory.
In practice when soaring across country, to find out what the speed of the wind is, you fly a square and plot your drift over a known mark.hence my use of the term ground speed. These days with a GPS it is much easier to plot ground speed versus indicated airspeed.
Non the less the object of the exercise, is to find a spot on the polar curve of the aircraft that gives you the greatest forward distance, for the least loss of altitude.

 

[aviat]

The answer is found in page 5-7 Figure 5-11 of FAA-H-8083-13 Glider Flying Handbook (2003 edition) and in the sites below. It is as berkshire says. You determine L/Dmax from the polar curve in still air, which should be in your GFM/ POH. This gives you the IAS (or CAS) to fly for that weight.From that you determine the effect of wind.

I understand how you come up with it being min drag but I don’t understand your statement about TAS. You fly by adjusting indicated or calibrated airspeed, for instance, increasing the weight changes the airspeed for L/Dmax, as does configuration changes like flaps. Each of these changes has a different polar chart.

A general rule of thumb is to add one half the headwind component to the best L/D for the maximum distance.

http://home.att.net/~jdburch/polar.htm.

http://www.auf.asn.au/emergencies/aircraft.html

 

[berkshire]

Aviat,
Thank you for pulling my irons out of the fire. I know I have a "Joy of Soaring" some where around here.
I was making the mistake of trying to answer this question without cracking a book.
Anyway your website " jdburch/polar" has all the information and the Mcready solution which covers most of what anybody could want to know about polar curves in rising sinking and moving air.
B.E.

 

[rb1957]

actually, don't you maximise range by maximising thermal up-drafts ? ... yes, i know that's the practical answer, and you're looking for the theoretical one, but ...

 

[berkshire]

Actually you do not maximise thermal up drafts, that is what the Mcready solution is about. he discovered that thermals are weak at the bottom and weak at the top.
Therefore it is better to leave the thermal at its strongest point and go find another one. Remember the object is to progress across country, not stay in one spot.
B.E.

 

[rb1957]

my point was that if you knew there was a strong thermal somewhere nearby then (i'd have thought) you'd pretty much rush (if you can rush in a glider) there to take advantage of it; rather than tooling on over to it at a theoretical optimal long range speed (by which time it might have dissipated.

 

[berkshire]

rb1957
That is precisely what you do. The Mcready solution gives you a speed ring, that is fitted to your rate of climb indicator, which tells you how fast you can go, based on the strength of the thermals you are encountering.
Check out the link Aviat supplied (jdburch / polar) it explains this much better than I can here without writing 2 pages.
B.E.

 

[MAAero]

There exists an analytical way to determine the speed which would maximize the ground range.

Assuming a drag polar of the usual form Cd=Cd0+kCl^2, and assuming the rate of descent to be small, the true airspeed, V, should be chosen to satisfy the following equation:

Vw/V1 = (2m(1-m^4)) / (3m^4-1)

Where Vw is the wind speed
V1 is the minimum drag speed with respect to a still atm.
m= V / V1

This can be proven by considering the glider flying in a "pocket" of still air, the pocket itself moving at a speed Vw. The ground range, in this case, is
R=(L/D)delta_h + Vw*delta_t

Where delta_h and delta_t are the differences in height and time. The proof gets too long to write it here, but for sure it can be done.

 

[mechgyver]

MAAero,
Is there any chance you could share that proof with me? Or at least direct me to where I can find it?
Thanks!

 

[khlama]

thank you very much all for your help that was really appreciated

many thanks again
I was able to prove MAAero relation ...
Some algebra is required, but you endup indeed getting this relation

Thanks again

With best regards

 

[mechgyver]

Same here. Not as convoluted as I had initially expected.