The glide angle is the acute angle the glider’s flight path makes with the horizontal floor. The angle can be measured by making a scale drawing of the gliders flight triangle. The horizontal leg of the triangle will represent the distance traveled over the floor and the vertical leg will represent how high the glider was above the floor when launched. This is best done with an engineer’s scale, having various divisions of an inch. With school rulers, it might be best to let 1/4″ on the drawing represent one foot of actual distance, or whatever it takes to fit the diagram onto a sheet of paper. Quadruled paper makes this easy.

If the glider was launched 4′ above the floor, that distance would be represented by a 4 x 1/4″ = 1″ vertical line on the scale drawing. If it flew 20 feet, that distance would be represented by a 20 x 1/4″ = 5″ horizontal line on the drawing. Now draw in the hypotenuse, representing the glider’s flight path, and use a protractor to measure the angle between the horizontal line and the hypotenuse, in this case 11 degrees. (This can also be calculated from the distances, since the ratio of the height to the distance is the tangent of the angle. In my school days, we would look this up in trig tables, but today scientific calculators are widely available and you can find the angle as the arctangent of 4’/20′ or 0.2, which works out to 11.31 degrees.) This is known as the glide angle. You can also measure the length of the hypotenuse and find the distance the glider traveled through the air, in this case 5.1 inch on the drawing, corresponding to 4 x 5.1 = 20.4 feet. This shows that the distance the glider flew through the air is a little greater than the distance it traveled over the ground. (This distance can also be calculated from the theorem of Pythagoras as the square root of the sum of the squares of the other two sides, easy to do with today’s scientific calculators.)

** Calculating Airspeed**

This allows estimation of the glider’s speed through the air. Divide the distance traveled through the air by the time of flight to get the glider’s speed through the air. Assume the time of flight was 3 seconds. The speed through the air is then 20.4 / 3 = 6.8 feet per second. You notice that this speed is different from the speed over the ground, because the distances are different.

Both of these differ from the descent rate for the same reason.

**Forces of Glide**

In a steady glide the weight of the glider pulls it downward along an inclined path and the air flowing over the glider produces an aerodynamic force that is equal and opposite to the weight.

** Lift to Drag Ratio**

The aerodynamic force may be resolved into two components. Lift is defined as the component that acts perpendicular to the flight path in an upward direction. Drag is defined as the component that acts parallel to the flight path opposite to the direction of flight. The lift to drag ratio is numerically the same as the ratio of the distance traveled horizontally to the height descended.

This is because the weight acts parallel to the height, the drag acts parallel with the flight path and the lift acts perpendicular to the flight path, tying these triangles together. In the hypothetical case here, it would be 20′ / 4′ = 5. This is important to a glider pilot because it tells how far the glider will go from a given altitude. It helps the pilot decide when to head home.

** Force Diagram**

This is just the ratio of the lift to the drag. It doesn’t tell us what the actual forces are. To do that we need something that establishes a scale of force. Lift and drag are forces. The weight of the glider is the force that establishes the scale of the diagram of forces, so a balance is required to weigh the glider.

The triangle of forces is geometrically similar to the glide triangle. You can see this by drawing the glide triangle, as above, and plotting the lift, drag and weight forces acting on a point along the hypotenuse. Pick any point about midway along the hypotenuse of the glide triangle. This point represents the glider’s position at some time. In our example we know that the drag force is one unit long for every five units of the lift force. Draw an arrow 1/4″ long pointing uphill along the glide path from the point. This represents the drag force. Draw an arrow perpendicular to the hypotenuse, pointing upwards, 5 times as long as the drag arrow, or 11/4″ long. This arrow represents the lift force. These two arrows form two sides of a rectangle. Carefully draw in the other two sides of that rectangle. Now draw in the diagonal of the rectangle that passes through the original point. This diagonal will be vertical.

This diagonal represents the total aerodynamic force that acts on the glider and it is equal and opposite to the weight force in a steady glide. The triangles formed by the diagonal within the rectangle are geometrically similar to the glide triangle. Both triangles have the same angles. Corresponding sides are in the same ratio to each other. In particular, the lift (long side of force triangle) is to the weight (hypotenuse of force triangle) as the horizontal distance covered (long side of flight path triangle) is to the flight path length (hypotenuse of flight path triangle). Let’s assume the glider weighs 10.2 grams.

If the 10.2 grams of the glider’s weight corresponds to the 5.1 inches of the scale drawing of the flight path, then the lift must be 10 grams to correspond to the 5 inches of the scale drawing of the horizontal distance covered. The ratio is two to one.

This can be expressed algebraically, and it can be expressed in terms of the original measured distances. We also know that the drag is one fifth of the lift, so it must be 2 grams. This establishes a scale for our force diagram of 2 grams per quarter inch or 8 grams per inch. The scale force diagram is known as a vector diagram. The arrows representing the forces are known as vectors and weight is the vector sum of lift and drag. A vector is a mathematical entity that has both magnitude and direction, just like forces.

** Gravitational Energy of Glider**

A glider uses gravitational energy to fly. Gravity doesn’t just pull the glider down, it also pulls it against air resistance. Gravity powers the glider. From the information we have, we can calculate the energy and power provided by gravity.

** Definition of Work**

When you lift a weight, you do work. When you lift twice as much weight, you do twice as much work. When you lift the weight twice as far, you do twice as much work. Work is energy. Energy has units of a force (weight) times a distance (height). The energy that the glider had at the top of its path, relative to the ground, is the weight of the glider times its height above the floor.

That’s how much energy it would take to lift it that high and that’s how much energy it takes for it to fly against air resistance on the way down.

**Energy Units and Conversion**

Energy may be expressed in different units, depending on what units were used to measure the weight and the distance.

In the example above we had a 10.2 gram glider launched 4 feet above the floor. This would result in an energy value of 10.2 x 4 = 40.8 gram feet. It is more usual to express energy in foot pounds. There are 453.59 grams in a pound, so we must divide 40.8 gram feet by 453.59 grams per pound to get 0.0899 foot pounds of energy. Gravity expended that much energy to bring the glider down to the floor.

**Calculating Horsepower**

Power is the rate at which energy is expended. We know how long it took for that 0.0899 foot pounds of energy to be expended because we timed the flight. Let’s assume the flight lasted 3 seconds. Then we know that energy was expended at a rate of 0.0889 foot pounds in 3 seconds, or 0.0899 / 3 = 0.0300 foot pounds per second. It is more usual to hear power expressed in horse power. One horse power is 550 foot pounds per second, so we must divide 0.0300 foot pounds per second by 550 foot pounds per second per horsepower to get 0.0000545 horsepower.

Gravity pulled the glider against air resistance. We measured the air resistance as 2 grams of drag. We also measured the distance the glider moved through the air against that drag, 20.4 feet. The amount of energy expended by gravity pulling the glider against the drag is the force times the distance, or 2 grams x 20.4 feet = 40.8 gram feet, the same as we got before.

This took the same 3 seconds, so the horsepower is also the same when calculated this way. The interesting thing about this way of calculating the power is that we multiplied the drag force by the length of the flight path and divided by the time of flight.

But the length of the flight path divided by the time of flight is the glider’s airspeed. This tells us that another way to calculate the power of a glider is to multiply its’ drag by its’ airspeed.

**Science and Math Lessons from Simple Gliders**

This illustrates how simple gliders can motivate a discussion of measurement, length, time, speed, unit conversion, ratio, angle, relative speed, piloting, scale, scale drawing, graphical solution, right triangles, angle tangent, Pythagorean theorem, weight, force, force diagram, vectors, vector addition, geometrical similarity, energy, rate and power. It also illustrates how scientific concepts begin with simple operational definitions in measurement processes and are elaborated through geometrical and algebraic manipulation. You can draw a flow diagram showing how the measured quantities lead to the derived quantities.

With four simple measuring tools, a tape, a protractor, a stopwatch and a balance, we have measured four quantities and derived ten. We have demonstrated numerical and graphical solution methods.

To Build Simple Gliders From Foam Plates & Straws check article on this website.

Hmm is anyone else encountering problems with the

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On this page there are only 3 drawings so it should load quick, on some of the other pages there are a lot of images so loading could be slower. I had used tables to position many of the images and IE 11 does not handle this the same as IE 10 which has been giving me fits.

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