# Pitch of a glider with altitude

I'm designing a high-altitude glider. Just to clarify a few points before my question:

• The glider will weigh close to 200 grams.
• It will be using the A18 airfoil.
• The glider is dropped from 80,000 feet.

Please correct me if any of my following statements are incorrect.

I'm calculating the pitch that my glider should be flying at by finding the glide angle. The glide angle does not change with altitude. This means that the pitch my glider should be flying at should not change either, so the elevator should not need to be used, and I will just set my CG to have the nose pitching down at the glide angle. However, the altitude does change the Reynolds number of the air. According to airfoiltools.com, the most efficient alpha changes with Re #, so for maximum efficiency, a glider will need to change the alpha. The most common way I know of changing the alpha is by using the elevator of the glider. This changes the alpha - but it also changes the pitch of the glider! So, the glide angle will change. What should I do about that? Is this the best way of calculating pitch?

• Isn’t the alpha the same as the angle of attack, therefore the same as the pitch? That’s what I’ve always thought. Also, the angle of attack that you’ll need for level flight will change with altitude. The air is thinner at higher altitudes, especially at 80,000 feet. The plane needs more AoA to keep the same lift because there is less air per volume. Apr 12 at 2:58
• Also I would assume because it’d be at such high altitudes, the AoA it would be forced into to retain level flight wouldn’t allow you to compensate for the most efficient AoA from Re number. Dependent on speed though Apr 12 at 3:09
• Pitch =/= AOA @Wyatt Apr 12 at 4:08
• @RobertDiGiovanni Is there a book you'd recommend? Apr 12 at 11:04
• Fascinated by the aerodynamics aspect of the question, but as a pilot I ought to mention you're going to need a transponder (in my local airspace at least...) and a ton of permissions to drop something from 80k feet with an arbitrary 100NM+wind range. You may want to re-think the 200g payload... Apr 12 at 12:38

The angle of the velocity vector relative to the ground reference frame is the flightpath angle, or the climb angle. Typically a $$\theta$$ or $$\gamma$$ depending on which reference you use.

The angle of a reference line on the aircraft to the velocity vector is the angle of attack. Universally $$\alpha$$.

The sum of these angles $$\theta+\alpha$$ is therefore the angle between the reference line on the aircraft to the ground -- this is what an observer on the ground sees as the airplane flies by. It is sometimes called the 'deck angle'.

At glide, the flight path angle is determined by the lift to drag ratio. You can draw some velocity diagrams and free body diagrams to convince yourself of this.

$$L/D=-1/\tan(\theta)$$

As discussed in the other thread, Best L/D will occur at a certain lift coefficient. Every other lift coefficient will also have a corresponding L/D. You can calculate the L/D for a given CL from the drag polar.

If you also have a curve for $$C_L$$ vs $$\alpha$$, then you can find the angle of attack corresponding to your $$C_L$$ and also your L/D.

Put it all together and you can calculate whatever you need, $$\alpha$$, $$\theta$$, and their sum, the deck angle.

"Pitch" can be confusing in aviation because we have "pitch to the horizon" and "pitch control" with the elevator.

For purposes of clarity let's separate "pitch" into glide ratio and Angle of Attack.

As seen at airfoil tools, the lift to drag ratio àt Re 500,000 will be much greater than 50,000.

Essentially, what this means is you get more lift for the same amount of drag.

In a stabilized glide lift requirement will be constant

You need to study static stability.

Static stability controls the aircraft lift production by controlling airspeed. No, not with a fancy computer, just by setting the elevator. Elevator controls angle of attack.

At a higher Re, the wing works better, therefor the indicated airspeed needed for adequate lift (for a given weight) will be less.

As always, when you release your glider, it will dive until (whatever) adequate airspeed is reached to allow it to glide in a straight line towards earth.

The angle of this glide path is determined by have much potential energy (altitude) is consumed per horizontal distance traveled. This is your glide ratio.

If you have a higher Lift to Drag ratio, Lift being constant, the plane will have less drag to produce the same amount of lift.

thrust = drag

Glider "thrust" is the sine of the angle of its flight path towards earth × weight

What this means is the glider will have a higher glide ratio at altitude. Again we must realize a glider dropped from 80,000 feet will easily glide more than 100 miles.

for any given elevator setting, the glide ratio will be higher at altitude

For this simple experiment, taken to the extreme of 80,000 feet, plan on greater gliding distance.