By Jim Davis    2021-04-25

# Helicopter Landing Gear

Helicopters may use wheels, floats, or skids. Each type of landing gear has pros/cons and the best one depends on the mission/application the helicopter will be used in. In this article, we’ll briefly discuss the various landing gear and the pros/cons of each.

If speed and range are a priority, retractable landing gear are favorable. Retracting the landing gear reduces drag allowing the helicopter to fly faster and further. On the downside, retractable landing gear are more complex, adding to the cost, maintenance and weight of the helicopter. For larger, heavier helicopters the weight/cost associated with a retractable landing gear is relatively small (compared to the total weight/cost) and hence may be more attractive.

Wheeled landing gear are more popular on larger/heavier helicopters. Wheels simplify the movement of helicopters around an airfield, allowing them to roll from location to location. Moving helicopters with skids often requires “hover taxiing”—taking off into a hover and slowly flying to a new location. It is possible to use other machines to lift/transport skid-equipped helicopters, but of course those machines have their own cost. Hover taxi operations are less practical for larger helicopters that produce more downwash.

Wheels add weight/cost relative to skids. However, the cost of adding wheels is relatively smaller for larger helicopters (compared to total helicopter weight/cost). As a general rule, wheels are used on helicopters weighing more than 4 tons.

Wheels may also improve run-on landings. There are situations (typically engine emergencies) where a helicopter won’t have enough power to land with 0 speed and a run-on landing (which requires less engine power) is required.

Wheeled gears can be made to swivel either passively or steered by pilot control. The former is simpler but less controllable than the latter. Again here the more costly/complex option is more likely to be used on a heavier, more expensive helicopter.

Skids are superior to wheels when it comes to simplicity and cost. There are other advantages to using skids. For example, ground resonance is less likely to occur with a skid. Helicopters that need to land in softer areas (e.g. snow or swampy areas) may be equipped with skids and add-on “bear paws” to prevent excess sinking into the terrain. In fact, this was the motivation behind the invention of skids by Bell Helicopter in the 1940s—reports of early wheeled helicopters getting stuck in muddy/swampy terrain.

Of course, skids and wheels are not useful for landing in the water. These helicopters use buoyant floats to stay atop the water. In addition, some helicopters have emergency floats that can inflate/deploy in the event of a water landing. Most of these are highly susceptible to overturning in high sea states.

## Math for a swiveling nose gear

Here we discuss turning on the ground mathematically. Specifically, we’ll treat a swiveling nose gear (passive or active) and relate the angle of this gear to the helicopter speed, yaw and turn radius. We start with a few definitions.

 NLG nose (front) landing gear LLG left (aft) landing gear RLG right (aft) landing gear $$A$$ (swivel) angle of the NLG tire relative to the aft gears $$u$$ forward velocity of RLG $$v$$ lateral velocity of RLG (typically assumed 0) $$x$$ longitudinal distance (along the stationline) from the NLG to the RLG (and LLG) $$y$$ lateral distance (along the buttline) from the NLG to the RLG $$r$$ yaw rate about the RLG, positive clockwise as viewed from above $$R$$ turn radius

We’ll consider turns to the right and base much of the math on the RLG. For simplicity, we’ll use the velocity $$u,v$$ and yaw rate $$r$$ about the RLG. Note this is not equivalent to the velocity and yaw rate about the CG. For a simple example, consider the case where the helicopter pivots about a stationary RLG.

We’ll first compute the velocities at the NLG and LLG given the values at the RLG. These are the velocity at the RLG plus the components due to the yaw rate $$r$$ about the RLG. We neglect pitch and roll motion since we’re on the ground, and we’ll assume $$v=0$$: the RLG (and LLG) are not skidding laterally. This gives the following equations. (Notice that $$v_{LLG}$$ has no component due to $$r$$ because it’s at the same stationline as the RLG.)

$$u_{NLG}=u+ry$$
$$v_{NLG}=v+rx=rx$$
$$u_{LLG}=u+r(2y)$$
$$v_{LLG}=v=0$$

We’d like to find the condition in which the NLG is not skidding laterally either. In this case, however, the wheel is swiveled by the angle $$A$$ relative to the aircraft, as shown in the diagram. In the NLGs “tire reference frame” we have $$u’_{NLG}=u_{NLG}\cos A+v_{NLG}\sin A$$, or $$u’_{NLG} = (u+ry)\cos A+rx\sin A$$, and similarly $$v’_{NLG} = rx\cos A- (u+ry)\sin A$$.

So if the NLG is not skidding laterally—$$v’_{NLG}=0$$—we get the equation $$rx\cos A = (u+ry)\sin A$$ or $$r = \frac{u+ry}{x}\tan A$$. We can also use this equation to derive the NLGs swivel angle from the $$u,r$$ to get $$A=\arctan\frac{rx}{u+ry}$$.

## Pivoting

Let’s check the math above in the case of pivoting on the RLG. Pivoting means the RLG remains fixed on the ground and the helicopter rotates about the RLG. In this case, you can see from the picture that $$A$$ should be perpendicular to the line between the RLG and the NLG, i.e. $$\tan A = x/y$$. You can verify that this is indeed the case, using the equation derived above with $$u=0$$ and canceling the $$r$$.

It’s sometimes useful to estimate the turn radius of a helicopter. This is the radius of the circle made when the helicopter does a full 360 deg turn on the ground. Note that the turn radius of different parts of the helicopter is different. When the helicopter pivots on the RLG, the RLG’s turn radius is 0, while the LLG’s turn radius is $$2y$$ and the NLG’s turn radius is $$\sqrt{x^2+y^2}$$.
We can find the turn radius of the RLG using the definition of it’s speed $$u$$. Since $$u$$ is the distance traveled per unit time, and the RLG travels a circular circumference of $$2\pi R$$ in time $$2\pi /r$$ (the time it takes to yaw by $$2\pi$$), we get $$u=\frac{2\pi R}{2\pi /r}= Rr$$ or $$R=\frac{u}{r}$$.
As you might guess, we can also compute the turn radius from the nose wheel angle $$A$$ (assuming again that no tires are skidding laterally). We had above $$\tan A=\frac{rx}{u+ry}$$. Dividing the numerator and denominator by $$r$$ and using the equation for $$R$$ gives $$\tan A = \frac{x}{R+y}$$ or $$R = \frac{x}{\tan A}- y$$.