In a previous article I have told you how to locate the wings of a plane in relation to the center of gravity, but have not told you how to find the center of gravity itself, a most necessary operation.
Of course, you can first build the fuselage with all items in it and then balance it over a sawhorse, but this is a costly way to solve so simple a problem and may lead to a lot of lastminute alterations.
To begin our calculations, we must know the weights of all items such as the propeller, engine, gas tank, oil and tank, landing gear, fuselage, tail surfaces, and tail skid. We should also know the center of gravity of each of these items within itself. The engine manufacturer can give the location of the C. G. of his product or we can find this point ourselves by experiment.
In Table No. 1 are listed all of the items, their weights and the distances of these items from a reference line called the "Datum Line." This datum line is usually taken at the front surface of the rear plate in the propeller hub but any other point will do as well. These distances, which are measured from the datum line, are called the "LeverArms" of the weight items.
TABLE No. 1. Tabulated Items of C.G. Calculation   1.  2.  3.  4.  5.  6.   Item  Weight  Vertical Arm  Vertical Moment  Horizontal Arm  Horizontal Moment   Engine and Propeller  180  lbs.  0  in.  0  in. lb.  8  in.  1440  in. lb.  Gas and Tank  70   7   490   36   2520   Oil and Tank  10   6   60   27   270   Pilot  170   3   510   64   10880   Chassis  37   28   1036   25   925   Fuselage  75   3   225   46   3450   Tail Surfaces  10   2   20   114   1140   Tail Skid  3   9   27   114   342    Totals  555  lbs.    1228  in. lb.   in.  20967  in. lb. 

Notice, that in Column No. 1, that all items are listed except the wings. We are to find the center of gravity first without the wings and then later place the wings on the fuselage in proper relation to the center of gravity. In Column No. 2 are the weights of the items. In Column No. 3 are the vertical leverarms, or the distance that the item is above or below the thrustline. Column No. 4, called the "Vertical Moment," is the result of multiplying Column No. 2 by Column No. 3. Column No. 5 is the Horizontal LeverArm of each item, measured back from the rear plate of the propeller hub. Column No. 6 is the result of multiplying Column No. 2 by Column No. 5.
To find the Horizontal location of the center of gravity, divide the total of Column No. 6 by the total of Column No. 2, which is: 20,967 inchpounds divided by 555 pounds. This gives us 373/4 inches back from the rear surface plate of the propeller hub.
To find the vertical location of the center of gravity, divide the total of Column No. 4 by the total of Column No. 2, which is: 1228 inchpounds divided by 555 pounds or 2.21 inches. The minus sign () means that the measurement is below the thrust line.
The items and measurements listed in Table No. 1, are taken from the vestpocket pursuit biplane called "The Knight Twister." In Figure No. 1, we have made a sketch of The Knight Twister in its level landing position. The center of gravity of each item is shown by a small dot. The horizontal measurements from the datum line to each item, as well as the vertical measurements, from the thrust line are also shown.
After the problem is worked out with results tabulated as in Table No. 1, the center of gravity of the ship is measured off on the drawing as in Figure No. 1. A vertical line is drawn from the center of gravity of the ship down to the ground line and from this point we show a measurement forward of 12 inches to the point of wheel contact with the ground. The angle formed by the vertical line, and a line from the center of gravity to the point of wheel contact with the ground, is 15 degrees. This angle can be from 12 to 17 degrees as shown in Figure No. 4.
The landing gear must be long enough to give the proper amount of propeller clearance, which is 10 inches when in the level landing position and to allow us to get at least 95 per cent of the maximum lift when in the three point landing position. The angle of maximum lift of some wings is 18 degrees, but that of the M6 wing used here is 20 degrees. Some designers place the landing gear so that when the ship is in the threepoint landing position, the wings are at the maximum lift angle. Others set the landing gear so the ship is at only 10 degrees. This small angle allows you to use a shorter landing gear, thereby saving a lot of parasitic resistance.
In the Knight Twister, the diameter of the propeller was the deciding factor in designing the landing gear. Sometimes the landinggear is referred to as the "Chassis," as in Table No. 1. The landinggear may need to be further forward when brakes are used, but the position shown has worked out very satisfactorily.
In Figure No. 4, two angles are shown  Angle (B) and Angle (A). Angle (B) is the angle formed by two lines that designate the ground lines at level landing and at threepoint landing. The angle (A) locates the landing gear in relation to the C. G. In Figure No. 2, the angle (B) is 10 degrees and in Figure No. 3, the angle (B) is 18 degrees. Maybe this Figure No. 3 has also given you a big laugh, but there is something else there besides a laugh.
We are told that the angle (B) should be sufficient to give an angle of at least 95 per cent of the maximum attack to the wings that will result in lift of whatever type of airfoil we use. For a Clark "Y" it would be 12 degrees, with an M6 would be 161/2 degrees and a U.S.A. 40 wingsection would require 131/2 degrees. If these wings were set on the fuselage at an angle of incidence to give the best liftdrag ratio, then the Clark "Y" would be set at 1 degree, the M6 set at 4 degrees and the U.S.A. 40 at 1 degree.
In Figure No. 2, we drew a wing with an angle of incidence of 4 degrees. Now if the required angle of attack to get 95 per cent of the maximum lift is 14 degrees, it means that the angle (B) will be 10 degrees, for 14 degrees it will be minus 4 degrees (angle of incidence) equals 10 degrees for angle (B). In Figure No. 3 we drew a wing with an angle of negative (4) degrees incidence. If this wing had to have an angle of attack of 14 degrees, angle (B) would have to be 18 degrees; for 14 degrees, plus 4 degrees (angle of incidence) equals 18 degrees for Angle (B). Figure No. 3 is an exaggeration, but shows what happens to the length of the landinggear struts when the angle (B) is increased.
Notice that in Figure No. 3, the Angle (A) is 15 degrees or the same as in Figure No. 2, but the wheels are further forward. That is, the distance measured from the vertical line dropped from the C. G. to the ground line to the point of wheel contact with the ground is more in Figure No. 3 than in Figure No. 2. Enlarging the angle (B) causes the wheel contact point to be moved forward, thereby making the landinggear struts longer.
Figures Nos. 567 are the front views of Figures Nos. 234. Figure No. 7 shows two angles  angle (C) and angle (D). Angle (C) should not be less than 25 degrees. This angle is formed by a line dropped vertically from the center of gravity to the ground line, and a line from the center of gravity to the point of wheel contact with the ground. This angle (C) can be larger.
Angle (D) which, is formed by the ground line and a line drawn from the wing tip to the point of wheel contact with the ground, should not be less than 6 degrees and the wing or elevator tips should not drag on the ground with ailerons or elevators fully depressed.
In Figures Nos. 56, we drew the landinggears with the Angle (C) equal 25 degrees which is the minimum angle allowed. Notice that in Figure No. 5, the spread between the wheel treads is much less than in Figure No. 6. The wider this spread the longer the landinggear struts. One point that we are trying to get over is; the shorter the landing gear struts, the smaller amount of power consumed to drag the chassis through the air.
Figure No. 5 shows a very narrow spread to the landinggear  most planes have it wider than this. Too wide a spread with a short fuselage will cause the ship to ground loop when one wheel strikes something on the ground. For a ship that is 20 feet overall length, the spread of the wheels could be 5feet without any undue annoyance from ground looping.
In Figure No. 4, angle (A) and in Figure No. 7, angle (C), are both drawn from the center of gravity. If the center of gravity is high, the wheels will touch the ground further ahead and further apart.
In Figures Nos. 234567 we used a long fuselage with a lowwing monoplane. With a short fuselage and a shortspan biplane, the landinggear may. not have such long struts, as pictured in Figure No. 3. See Figure No. 1, the side view of the Knight Twister which has an angle (B) of 20 degrees. This angle can be smaller with a smaller diameter propeller.
If you prefer a wirebraced chassis instead of a monostrut type, it could be designed to have less resistance for the large monostrut does cause quite a large ripple in the slipstream. Of course, a retractable landinggear solves a lot of these difficulties with wind resistance, but it is hard to find room in a small fuselage.
We see that certain items are entirely dependent upon personal judgment and preferences. If it were not for this condition, there would be no point to start from. After choosing the type of wing bracing, type of landing gear and the type of fuselage, we then move things around until they all match up according to the rules given here.
After the C. G. and landing gear details are adopted, then we must check up locations with the requirements for the propeller clearances. This means that we must go back to the power plant and work out the probable propeller diameter before we can safely adopt any given height of landing gear.
The propellers in a twoengine job should clear the fuselage by 6 inches. The propeller on a seaplane should clear the water by at least 18 inches and clear the floats by not less than 2 inches.
