Landing Gear Loads

To my knowledge, the calculation of landing gear loads for gyroplanes has never been documented. My following article is based on my calculations of landing gear loads for fixed wing aircraft and the Code of Federal Regulations (CRF) Part 23.

The 14 Code of Federal Regulation (CFR) Part 23 states that the limit inertia load factor for the design of aircraft landing gears must be determined by a drop test or by some rational means. Since a drop test is not possible during the design phase we must look to a rational means to design the landing gears. This analysis is that means.

The main landing gears on a gyroplane may be viewed by looking at the aircraft from the front as shown in figure 1. The shock absorption of the landing shock occurs just as the tires touch the ground and continues as the tire and landing gear leg deflect and absorb the energy of the vertical sink velocity, Vs, of the landing gyroplane. A simplified view of the two conditions of landing are shown in figure 2. Each condition in figure 2 is the same view as figure 1 but the landing gear and tires have been represented by a single spring with a spring constant of K in lb/inch or in lb/feet. To convert simply multiply lb/in. by 12 to get lb/ft.

According to Paragraph 23.473 (d) of Part 23, the velocity Vs depends on the wing loading which we will term the rotor disc loading, Wt/S for the gyroplane at landing weight as shown in equation 1.

Vs = 4.4(Wt/S)1/4                                                                                                                         (1)

Vs is in feet per second
Wt is the landing weight of the aircraft in pounds
S is the rotor area in square feet

For fixed wing aircraft weighing less than 12,600 lbs, Part 23 also states that when Wt/S is equal to or smaller than 6.4 psf, Vs is equal to and cannot be smaller than 7 fps. However, gyroplanes have extremely low disc loadings and land softly at low Vs. I therefor propose that we use equation 1 to determine Vs. The free fall height, h, in inches to obtain Vs are given in equation 2.

h = 12 Vs2 / (2 x g)                                                                                                                         (2)


Vs is in fps
g is the acceleration of one gravity equal to 32.2 feet/second2

Now if we look at condition 1 in figure 2, when the wheels are just beginning to touch the ground at time = 0, the kinetic energy is,

K.E. = Wt Vs2/(2 g)                                                                                                                         (3)

Paragraph 23.473 (b)(1), Part 23 allows the landing weight, Wt, to equal 0.95 x gross weight, W.

  Figure 1. A View Aft of the Gyroplane showing the Landing Gear Dimensions and Forces.
  Figure 2. Two Conditions For the Landing Gear Stroke. Same as Figure 1 except Landing Gears are shown by a Spring.
  Figure 3. Testing the Tire Deflection under a Load of up to 6,500 lbs.
6.00 x 6 Tire                                  5.00 x 5 Tire
Table 1. The Load/Deflection of a 6.00 x 6 Tire with 6 plies and a 5.00 x 5 Tire with 6 plies for Various Tire Pressures.

During the landing shock, we are allowed to account for an amount of lift from the wings (rotor) since the aircraft is at a speed just below stall and high angle of attack (nose up). The lift coefficient for the stalled wing is usually taken to be from 0.4 to 0.5. If we express the wing lift as a lift factor, P equal to the wing lift divided by the aircraft landing weight, Wt, then the lift during landing is,

Lift = Wt (1 – P)                                                                                                                              (4)

For a wing lift coefficient of 0.4 to 0.5, it can be shown that P = 0.667, However for a non feathering rotor that cannot stall, the rotor lift during landing is almost equal to the flight lift and the value of P is set equal to 1. The potential energy, P.E., during the landing gear stroke, X, is,

P.E. = Wt (1 – P) X = 0                                                                                                                    (5)

At the bottom of the landing gear stroke the vertical velocity, Vs = 0 and K.E. = 0 and the force that the tires and landing gear leg exert upward on the aircraft is,

F = X K                                                                                                                                             (6)

For a linear spring constant K, the energy that has been absorbed by the landing gear is,

E = ò Fdx = Kò x dx = K X2/2                                                                                                              (7)

This equation is accurate since the tire, as seen in figure 3, and the spring gear leg has a linear spring constant. If an oleo strut is used for the landing gear leg, the reaction force, R, of the landing gear is constant such that,

E = 0.9 R ò dx = 0.9 R X                                                                                                                    (8)

Now from the conservation of energy, the energy of condition 1 is set equal to the energy at condition 2 in figure 2.

K.E. + P.E. = E                                                                                                                                  (9)

Substituting equations 3, 5 and 7 into 9 we have,

K X2/2 – Wt Vs2/(2 g) = 0                                                                                                                 (10)

From which,

X= (Wt Vs2/(K g))1/2                                                                                                                         (11)

To solve for X, the only unknown in equation 11 is the spring constant K for the landing gear. K is made up of two spring constants. One is the deflection of the tire under load and the other is the deflection of the gear leg or strut.

I performed tests on a 6.00 x 6 tire with different tire pressures as shown in figure 3 with the results plotted in table 1. For a tire pressure of 40 psi a maximum tire deflection of 3.6 inches is realized when the tire is loaded to 5,500 lbs. The tire spring constant, ktire is 5,500/3.6 = 1,528 lb/in = 18,333 lb/ft.

From the cantilevered beam equation, (see reference 3) the spring constant of the gear leg in figure 2 can be calculated as follows. The deflection of the gear leg under load R is,

y = R z3/(3 EI)                                                                                                                                 (12)

For the gear leg section of figure 2, the section moment of inertia,

I = t3 w/12 = 13 6/12 = 0.5 in.4 and for aluminum the modulus of elasticity is,
E = 10 106 psi
z = 25 inches and let R = 1 lb, then from eq. 12
y = 1 x 253/(3 10 x 106 x 0.5) = 0.00104 in.
kleg = 1/y = 961 lb/in. = 11,538 lb/ft.

Since the tire and gear leg are in series and there are two such systems, the system spring constant for both the tire and gear leg is,

K = 2 [ktire x kleg/(ktire + kleg)]                                                                                                        (13)

Substituting into equation 13 gives,

K = 2 [18,333 x 11,538/(18,333 + 11,538)] = 14,163 lb/ft.

Part 23 allows the landing weight, Wt to equal 0.95 x gross weight. For a gross weight of 1,232 lbs the landing weight is 1,170 lbs. The gyroplane has a disc diameter of 30 ft and hence a disc area of p r2 = 3.141 x 152 = 707 sq.ft. so that Wt/S = 1.74 lb/sq.ft. Vs is determined from equation 1 as, Vs= 5 fps. From equation 2, we have a drop height of 4.7 inches. Substituting into equation 10, gives a gear deflection of,

X = [1,170×52/(14,163 x 32.2)]1/2= 0.25 ft. or 3.04 in.

Now multiply the deflection, X by the stiffness, K, and dividing by the landing weight, we can determine the limit ground load factor, ng,

ng = (X K)/Wt                                                                                                                                (14)

Substituting into equation 14, ng for the example gyroplane is,

ng = (0.25 x 14,163 )/1,170 = 3.03 gs

The landing gear of the Sportster was designed to 3.0 gs and many hard landings in a period of 29 years have proven that this load factor is adequate. No landing gear or other structural failures have ever occurred in the Sportster over this time period. The landing gear for the Bumble Bee are designed to 2.5 gs. It should be noted that ng must be multiplied by a safety factor of 1.5 for ultimate (failure) load.

I also need to point out that if the gyroplane has been built, it is easy to determine the landing gear stiffness, K, from weight and balance data and a simple wheel deflection test. Let us say that the load on each main wheel is Wm pounds. With the gyroplane empty, the weight on the main wheels is known. The distance of the fuselage above the ground is measured. With a hoist connected to the airframe, the gyroplane is lift so that the wheels just touch the ground and the new distance of the height of the fuselage off the ground is measured. The difference between the readings is u feet. K is now simply,

K = 2 Wm/u                                                                                                                                  (15)

After going through the above calculations, we realize that for the calculations of many iterations in which we try to optimize lets say the tire pressure or landing gear design, it would be best to use a computer program. Such a program is written in BASIC as shown in the next pages.
The relation between the gear load, ng, and the aircraft load, n, with P=1.0 is shown in Figure 4.

Figure 4. The aircraft load factor n is acting on the people and equipment inside the aircraft and ng/2 is acting on each landing gear leg as shown above.


  1. 14 CFR, Part 23, “Airworthiness Standards: Normal, Utility, Acrobatic, and Commuter Category Airplanes.” U.S. Government Printing Office, Washington D.C. 1992.
  2. Ladislao Pazmany, “Landing Gear Design for Light Aircraft, Vol. 1,” Pazmany Aircraft Corp., Box 80051, San Diego, CA 92138. 1986.
  3. Raymond Roark, “Formulas for Stress and Strain,” McGraw-Hill Book Co. 1965.