Glider question: best speed to fly

rottydaddy

rottydaddy

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Doing some review of my study materials, and looking at the stuff about polars, with wind or sink, I realize I don't quite understand something...
Having not really done any xc (and without a McC ring), I haven't had the chance to really study the matter empirically.... but I totally get flying faster than best L/D in sink to get out of it ASAP, sacrificing a little altitude to avoid losing a lot of altitude.

But I don't get this: if distance is the goal, how does it help to fly faster than best L/D, in sink or a headwind, if any speed other than best L/D will always yield a less favorable glide ratio, and higher rate of descent?

Obviously, it pays to abandon min sink speed in sinking air, esp. if you want to scoot over to that field or runway before you get too low... but going above best L/D isn't going to get you more distance , especially in a headwind.
So why is it recommended, for sink or wind? Is it because in practice, it usually works out, because sink is a bigger problem than headwinds for meeting distance goals?

As Vinnie Barbarino used to say: "I'm so confused..."
 
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Best l/ d applies in a no wind situation. Min sink applies mostly in rising air into a headwind or is assumed to be a good replacement for best l/d (best glide) with a good tailwind. Conditions exist where it is best to fly anywhere between or beyond them depending on wind, direction of flight, and proximity to rising or sinking air. Gliders are slick and going above max l/d keeps you in the headwind for less TIME. The same concept applies to powered flight but best practices dictate its better to assume best glide in a powered aircraft when the engine is lost as it is considered an emergency. In a glider it is not.

Time is the factor here. This is why the faster an aircraft cruises, the less a headwind affects its efficiency. An aircraft flying roundtrip. Into a 20 kt headwind, returning with a 20 knot tailwind takes longer for the trip than it would in a no- wind situation. Also a faster aircraft flying the same route will be affected less in their time to complete the same flight because they wasted less TIME in the headwind. At some point it becomes obvious if you imagine flying into a headwind that equals your airspeed. In this case, the only way to progress across the ground is to speed up by decreasing pitch and increasing airspeed above that of the moving air mass in which you are in.

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It helps to think of an extreme case to illustrate this. Consider both your headwind and L/D are 40 knots ...make sense?
 
Much more concise than mine. I applaud. I'll shut up now. Lol

Good example.

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It's because flying at best L/D gets you the furthest through the air mass for the least height lost, but that is not the same thing as getting the furthest over the ground. e.g. consider the trivial example of a glider with best L/D = 22 at 55 kts, flying through a 55 kt headwind. If you maintain best L/D, you will just sink to the ground without getting anywhere. So the optimal speed in a headwind is faster than best L/D. Sinking air is basically equivalent to a headwind - in either case it is robbing you of energy. Similarly, in a tailwind or with rising air the optimal speed is slower than best L/D.

What the actual best speed to fly is depends on the strength of the headwind component, the climb or sink rate, and the polar of the glider. Hence those "speed-to-fly" doohickeys.*

The same principles apply to powered flight if one wants to make efficient use of fuel, with the added variable of throttle setting.

* Googling reveals some statements that "speed to fly is independent of wind". I don't fully understand the reasoning behind that - perhaps it's the case when you're trying to fly a closed course, like gliding tasks tend to be, or perhaps it's to do with circling in thermals rather than optimizing a cross-country cruise speed for a single straight leg.
 
Glider Flying Handbook p 5-6.
"The best speed-to-fly in a headwind is easily determined
from the polar. To do this, shift the origin to the
right along the horizontal axis by the speed of the
headwind and draw a new tangent line to the polar.
From the new tangent point, draw a vertical line to
read the best speed-to-fly. An example for a 20 knot
headwind is shown in Figure 5-11."

It helps to see it illustrated in Figure 5-11, page 5-7
http://www.faa.gov/regulations_poli...s/aircraft/glider_handbook/media/gfh_ch05.pdf

BTW, the paragraph prior to this has an error in determining the glide ratio at best L/D . They chose the wrong point on the polar. It should be 50/2.2 which yields 23:1, not 26:1..
 
It's not distance but Speed. That is distance over time.

When lift is strong, you are willing to fly faster, and sink faster between thermals because you can climb faster in strong thermals. And that's independent of wind.

Having said that, almost everyone flys too fast because they don't account for risk, the risk of not finding that next strong thermal.

So load it down with water and get that lead sled moving!
 
I can't say I disagree with any of the above, and who am I to argue with a polar, but... if best L/D gives you, say, 20 miles from 5000 feet in still air, and best L/D plus half the headwind component gives you only 18 miles from 5000 feet (and the air mass is moving in the opposite direction), it doesn't seem right to fly faster than best L/D.

The "headwind equals airspeed" scenario seems to disprove that, duh. But... The glide ratio triangle from 5000 has a 20-mile base at best L/D. The whole triangle is drifting downwind; I get that. If it's drifting downwind at a speed equal to your airspeed, common sense says you have to go faster than the wind, or you won't cover any ground at all. Again, any idiot can see this. It's true even in cruise with power; airspeed is airspeed.

But if you fly faster in a glide, that base gets shorter. The faster you glide, even in a power plane with minimal or no power, the steeper your glide angle has to be. So the horizontal distance per vertical distance has to be less, right? How does a glide triangle with a shorter base get you farther, even if flying at best L/D will have you descending practically straight down (relative to the surface)? I get it, yet I don't... :mad2:
 
Bill Watson said:
It's not distance but Speed. That is distance over time.
Click to expand...
I'm speaking strictly about trying to cover a given distance in a headwind. Best L/D at first glance seems to give you the best chance, but obviously if the headwind component equals or exceeds best L/D, you won't cover any of that distance.
I accept it, but I don't understand why, if the glider will always go (through the air) farther, per foot of altitude, at best L/D. :dunno:
 
I think this example will best answer the OP's question.

You just struggled up in a 1 kt thermal to the altitude at which you can just make the airport at a McCready of 1 and you head for the airport. a Few minutes later you hit a 6 kt thermal so you have choice. You can keep gliding to the airport at an MC of 1 or stop and climb and then fly to the airport at a McCready of 6 to get the airport the fastest (ie in minimum time).

It is fairly obvious** that it is better(faster) to climb to an altitude at which you can glide to the airport at a MC of 6. Since it is a 6 kt thermal it is faster to stop,climb and glide to the airport at the faster speed than it is to continue at a minimum MC setting. This of course requires a fairly good computer that will tell you what altitude you need to get to the airport at the different McCready settings, you then need to leave the 6kt thermal at the altitude required make the airport at the MC of 6.

Also this gives you a better margin as if you do encounter some sink you can slow down once past the sink to still make the airport at lower MC setting.

Brian
CFIIG/ASEL HP16T

**Maybe it isn't that obvious, but that is one way to define McCready numbers. Essentially it means that if you encounter lift higher than the McCready setting you are currently using, then it is faster to climb and cruise faster than it is to continue at your current McCready setting. That is why if you are trying to fly at a McCready of 4 you should reject and leave thermals when they less than 4 and stay or climb if it is stronger than 4.
 
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brcase said:
I think this example will best answer the OP's question.

You just struggled up in a 1 kt thermal to the altitude at which you can just make the airport at a McCready of 1 and you head for the airport. a Few minutes later you hit a 6 kt thermal so you have choice. You can keep gliding to the airport at an MC of 1 or stop and climb and then fly to the airport at a McCready of 6 to get the airport the fastest (ie in minimum time).

It is fairly obvious** that it is better(faster) to climb to an altitude at which you can glide to the airport at a MC of 6. Since it is a 6 kt thermal it is faster to stop,climb and glide to the airport at the faster speed than it is to continue at a minimum MC setting. This of course requires a fairly good computer that will tell you what altitude you need to get to the airport at the different McCready settings, you then need to leave the 6kt thermal at the altitude required make the airport at the MC of 6.

Also this gives you a better margin as if you do encounter some sink you can slow down once past the sink to still make the airport at lower MC setting.

Brian
CFIIG/ASEL HP16T

**Maybe it isn't that obvious, but that is one way to define McCready numbers. Essentially it means that if you encounter lift higher than the McCready setting you are currently using, then it is faster to climb and cruise faster than it is to continue at your current McCready setting. That is why if you are trying to fly at a McCready of 4 you should reject and leave thermals when they less than 4 and stay or climb if it is stronger than 4.
Click to expand...
Yes, this is all clear to me. I haven't used a MC yet, but I get the primary purpose: to compensate efficiently for possible variables in lift and sink.
But again, I'm just talking about polars that recommend speeds higher than best L/D to compensate for a headwind when trying to cover a given distance. I understand why, in practice, you'd just want to fly however you need to in order to gain ground towards your goal, but I can't reconcile the practice with what I think I understand about best L/D and its relationship to distance. :dunno:
 
rottydaddy said:
Yes, this is all clear to me. I haven't used a MC yet, but I get the primary purpose: to compensate efficiently for possible variables in lift and sink.
But again, I'm just talking about polars that recommend speeds higher than best L/D to compensate for a headwind when trying to cover a given distance. I understand why, in practice, you'd just want to fly however you need to in order to gain ground towards your goal, but I can't reconcile the practice with what I think I understand about best L/D and its relationship to distance. :dunno:
Click to expand...

Maybe I didn't really answer your question, as I read it again.
Some others did post to pretty good answers, but the best answer is it might be time to get your glider polar and E6B out and try some examples with either constant head wind or constant sink value. Once you know the descent rate at a give speed you know how long it is going to take you to get to the ground. so then figure out how far you can fly. Then try it at a higher speed and the higher sink rate of the glider that goes with it and see how it works out.

Brian
 
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