Radial power

Hello everyone,

a few days ago I went to the flying field and a buddy brought a nice EDF Su-27 (don't remember the brand, so don't ask ). Anyway I noticed that the retractable landing gear was simplified in its operation, in that the main wheels went straight up, without rotating as in the fullsize. This got me thinking: how exactly does the fullsize retraction geometry work? I knew it involved a rotation around an oblique axis, but how does one find such an axis? I don't think the engineers at Sukhoi went by trial and error to find the correct angles, so there had to be a mathematical way to do it. And I wanted to find it

To get an idea of what I mean, take a look at these videos:


So there you have it, the result of quite a bit of brain-smoke coming out of my ears. Perhaps this can come in handy for someone building an Su-27 or Mig-29, as they both use a very similar main landing gear retraction scheme.


The problem

Let's start from the problem statement. The problem to be solved is designing the main landing gear retraction geometry so that:
1) when extended the main wheels are aligned with the longitudinal axis of the airplane (we'll not consider toe in or toe out for now);
2) when retracted the main wheels lie in a plane that is parallel to the ground plane, that is their axis is vertical;
3) the aforementined requirements must be accomplished with a single rotation movement around a fixed axis.

Now, as you can see in the videos the main gear does not rotate around an "obvious" axis, but it rotates around an oblique axis. This can be seen from the fact that in the videos the gear leg does not move in a single plane, but it temporarily points out while the gear is being cycled.

To find out how to angle this axis, we'll start from the simplest case and then we'll try and complicate it later.


The simple case: vertical gear legs when extended, no toe in or other strange stuff to account for

As the title says, in addition to the above constraints let's add these:

4) when extended, the landing gear leg is perfectly vertical (pointing straight down);
5) when retracted, the landing gear leg il perfectly horizontal (pointing straight ahead).

This situation is sketched in the first attached image. This shows the right main landing gear leg in both the extended and retracted positions, with a sketched wheel attached to the outside of the leg, as is the case with both the Su-27 and the Mig-29. As you can see, this follows the constraints that were given above: the leg is vertical when extended and horizontal when retracted; the wheel is parallel to the longitudinal axis of the airplane when extended and is parallel to the ground when retracted.
As a side note, all pircures were generated using Blender ( https://www.blender.org/ ) and then modified using MS Paint. Highly professional design tools at work here Notice that in Blender I disabled the perspective correction, so everything is shown with a parallel perspective.

In order to get to the solution, we need to give some names to some interesting points of this image. Take a look at the second picture, which is the same as the first picture with some annotations added.

What did I annotate, and why? Well, for starters I added a name O to the center of rotation. Here I made the assumption that the oblique axis of rotation does intersect the center of the main landing gear leg. Therefore I called this intersection point O.
Then, choose an arbitrary distance L (strictly greater than 0) and let's call A the point along the extended landing gear leg that is at a distance L from O. I didn't mark the distance L in the drawing, but OA = L.
Similarly, let's call A' the point along the retracted gear leg that is at distance L from O (OA' = L). We'll talk about B later.

Now an observation: the oblique rotation axis we are trying to find must respect the criterion that each point along it length must be at equal distances from A and A'. This is because in a rotation each point remains at a fixed distance from the rotation axis. So in order for the oblique axis to be correctly positioned to get the desired result, it must respect the equal distance condition.
But the locus af all points in 3D that are equally distant from 2 fixed points is a plane. And since O is also equidistant from A and A', then O must lie on this plane. So we have that our unknown axis must intersect O and must lie on the plane that is equidistant from A and A'. For now let's call this the oblique axis plane.

Now take a look at the third picture, where I tried to show a copule of planes (not airplanes ). The plane that is seen almost edge-on is the one I just talked about, the oblique axis plane. It is the locus of all points that are equidistant from A and A', and this plane also includes O. I also added another plane, the one that contains O, A and A'.

Now let's add another point B so that:
1) B lies on both the aforementioned planes (that is it lies on the line that defines the intersection between the planes);
2) the OAB angle is a right angle.

This generates a right triangle OAB which (by construction) lies on the same plane as AOA'. But it's also easy to see that OA'B is also a right triangle: we know that OA = OA' = L and that the OB side is shared. Plus, since both O and B lie along the intersection of the two planes (since they both belong to both planes), it's easy to see that the AOB and A'OB angles must be equal. This is a special case of a more general statement that says that every point X on a plane that is equidistant from two points A and A' will form the same angle between the plane and both the A and A' point (does this theorem have a name?). So, since AOB = A'OB, let's call this angle α (alpha).
Now we know that OA = OA' by construction, OB is shared between the two triangles OAB and OA'B and the angles AOB = A'OB = α. We can apply one of the triangle congruence criteria (SAS, see here https://www.mathsisfun.com/geometry/...t-finding.html ) and state that the triangles AOB and A'OB are conguent. This, in turn, means that AB = A'B and that OA'B = OAB, thus OA'B is also a right angle.

Take a look at pictures 5 and 6. consider a line passing through B and perpendicular to the AOA' plane and consider a point C along this line. We know that AB = A'B, that the BC segment is shared and that ABC and A'BC are both right angles by construction. Therefore the triangles ABC and A'BC are congruent, and so AC = A'C. But this meets the definition our oblique axis plane (the locus of all points that are equidistant from A and A'). Therefore C lies on this plane too.
At this point we can add another label for the BOC angle, which we call β.
You can take another look at this construction with pictures 7 and 8, showing everything from the top. Please notice that in picture 8 I highlight the fact that the angle shown there is not the true β, it's just a projection of β. So, even if in picture 8 it may look like the angle β is exactly 45°, that is just a projection of the true value of β, not it's true value. Be careful about this. Picture 6 shows the correct way to measure β, that is along the oblique axis plane (which we can now call the OBC plane).

Back to the main objective of all this, our goal is to find the angles α and β so that the OC segment defines our unknown rotation axis. But we are still missing one constraint: there is nothing to limit the rotation of the landing gear leg, so that our wheel may and up in any position.

To solve this there is one last step. Consider picture 9 where two lines were added passing through A and A' and parallel to the wheel axle in each position. The line passing through A is perpendicular to the AOA' plane, while the line passing through A' lies along the A'B line. Now let's call the angle formed between these two lines and the AC and A'C segments respectively γ (gamma) and γ'. If you think about it for a second, you can consider the AC and A'C segments as "handles" that rotate the landing gear leg as it is retracted. This means that the γ angle is the angle between the "handle" and the wheel axle. Same for the γ' angle. Since this angle cannot change as the gear is moved (unless the gear leg is twisted), then we have that γ must be equal to γ'.

We already know that the ABC and A'BC triangles are congruent and that the ABC and A'BC angles are right angles. So we have that γ' = BA'C = BAC. The line that defines γ is parallel to the BC segment (as they are both perpendicular to the AOA' plane), so it follows that γ = ACB.
The total sum of the internal angles of a triangle is always 180°, so since ABC = 90°, then BAC + ACB + 90° = 180° and so BAC + ACB = 90°. We already know that γ' = BAC, that γ = ACB and so γ + γ' = 90°. Since we also want γ = γ', then it follows that γ + γ' = γ + γ = 2γ = 90° and so γ = 45°.

But since γ = 45°, then it follows that BAC = ACB = BA'C = A'CB = 45°, and so AB = BC = A'B.

We are getting there. α is easy to compute, as it is half of the retraction angle. Since we assume that the gear leg is vertical when extended and horizontal when retracted, then the overall retraction angle is 90°. So α is half that, α = 45°.
As for β, we have to use a little trigonometry. Remember that we set OA = L. We can compute OB as: OB = L / cos(α). Also, we have that AB = L * tan(α). Then β can be computed from the tangent of the BOC angle, which is given by BC / OB:
β = atan(BC / OB)
= atan(L * tan(α) / (L / cos(α)))
= atan(L * tan(α) * cos(α) / L)
(since we chose L != 0)
= atan(tan(α) * cos(α))
(since tan(α) = sin(α) / cos(α))
= atan(sin(α) * cos(α) / cos(α))
(cos(α) != 0 since α != 90°)
= atan(sin(α))
(for α = 45°)
= atan(sqrt(2) / 2)
= 35.26°

So, for the simple case described above we have:
α = 45°
β = 35.26°

Finally, to confirm that all of this is correct, I performed a rotation around the computed axis, and it all matches perfectly (see the 10th picture).


Adding complexity: gear leg not vertical when extended

I know what you are saying: "hey, but the gear legs are not perfectly vertical when extended!". I know. But the above calculation can be applied almost exactly without modifications to the case of a non-vertical gear leg. Think about it for a second: if you rotate everything, then you can transform the case of a non-vertical extended gear leg into a case of a vertical extended leg with a non-horizontal retracted leg.

Suppose that the gear leg, rather than being vertical when extended, points forward with an angle ρ (rho). Refer to picture 11 and 12. Here I set an angle ρ = 12° (randomly chosen, just to pick a value). We can apply the above formula with the only caveat that now α is no longer 90° / 2, but it is given by:

α = (90 - ρ) / 2

Then β can be computed using the formula:
β = atan(sin(α))

So for ρ = 12° I get:
α = 39°
β = 32.18°

See picture 13 for the result.

Coming soon: toe-in

Radial power

Accounting for toe-in / toe-out

Ok, what if one wants to give some toe-in or toe-out to the main wheels when extended? Without Accounting for this, any toe in (or toe out) would remain when the wheel is retracted, so the wheel would no longer sit horizontal in its well. One would need to compensate for that somehow to make sure that the wheel returns to the horizontal position when retracted. Another possibility is that one may intentionally want the wheel NOT to sit horizontal in the wells, even if there is no toe-in or toe-out. Or maybe a combination of both requirements.

Edit: added two new pictures to better explain this part.
Take a look at pictures 3 and 4. Here we have a toe-out angle of 10°. As you can see γ is still measured between the AC segment and a line parallel to the wheel axle. But since we now have some toe-out, now the line parallel to the wheel axle is no longer perpendicular to the AOA' plane. So let's add another line passing through A and perpendicular to this plane. The angle between this new line and the one parallel to the wheel axle is obviously τ, that is the toe-out angle. Since we want the wheel to lie flat in its well, nothing changes for γ'.

We still have the constraint of γ = γ'. And we still know that since the triangles ABC and A'BC are congruent, then γ' = BA'C = BAC. But the since the line used to define τ is perpendicular to AB, then we have that BAC + γ + τ = 90°. But BAC = γ' and we want γ' = γ, so
γ + γ + τ = 90°
2γ = 90° - τ
γ = (90° - τ) / 2

This means that now γ may no longer be equal to 45°, so we can no longer have that BC is equal to AB and A'B. In this case we may have BC that is a different length.

We still have:
OA = OA' = L
AB = A'B = L * tan(α)
OB = L / cos(α)
BC = OB * tan(β) = L * tan(β) / cos(α)

The way we solve this is by computing γ with an arctangent and then finding β as a function of the other variables. Actually, since we know that γ = γ', it's easier to compute γ':
γ = γ'
= atan(BC / A'B)
(since A'B = AB)
atan(BC / AB)
= atan((L * tan(β) / cos(α)) / (L * tan(α)))
(We know that L != 0)
= atan((tan(β) / cos(α)) / tan(α))
= atan(tan(β) / (cos(α) * tan(α)))
(since tan(α) = sin(α) / cos(α))
= atan(tan(β) / (cos(α) * sin(α) / cos(α)))
(cos(α) != 0 since α != 90°)
= atan(tan(β) / sin(α))

So:
atan(tan(β) / sin(α)) = (90° - τ) / 2

Compute the tangent of each side:

tan(atan(tan(β) / sin(α))) = tan((90° - τ) / 2)
(since tan(atan(x)) = x)
tan(β) / sin(α) = tan((90° - τ) / 2)
(sin(α) != 0 since α != 0)
tan(β) = sin(α) * tan((90° - τ) / 2)
(arctangent on both sides)
atan(tan(β)) = atan(sin(α) * tan((90° - τ) / 2))

β = atan(sin(α) * tan((90° - τ) / 2))

Notice that when τ this formula reduces to:

β = atan(sin(α) * tan((90° - 0) / 2))
= atan(sin(α) * tan(90° / 2))
= atan(sin(α) * tan(45°))
= atan(sin(α) * 1)
= atan(sin(α))

Which was the formula that was derived earlier.

Notice that we talked about toe out, but it is enough to set τ to a negative value to have toe-in. For example τ = -2° represents 2 degrees of toe-in, while τ = 3° indicates 3 degrees of toe-out.

This can also be used to make the wheel sit at an angle when retracted, and you still use the τ parameter. And this is the reason why I chose to use τ to indicate toe-out: when the wheel has some toe out, then the retraction movement must rotate the gear leg more than normal, while when the gear has toe.in, then the movement must rotate the leg less than usual. Indeed positive values of τ are used for toe out (more rotation than usual), while negative values are used for toe-in (less rotation than usual).
So τ can be regarded as a "overrotation" parameter: τ = 5° means that the gear should rotate 5 degrees more than the "canonical" amount while retracting, while τ = -1° means that i degree less than "canonical" is desired.

With this, it should be fairly easy to understand how to use τ to account for both toe-in / toe-out and to place whe wheel in its well at an angle. A practical example: say one wants 2 degrees of toe-in and also wants to place the wheels in their wells at an angle of 5° with the outeboard edge of the wheel lower than the inboard edge. This latter condition is an underrotation condition, so the value of 5° becomes a negative value. How does one combine these requirements? Simple, do an algebraic sum:

τ1 = -2° (account for 2° of toe-in)
τ2 = -5° (leave the wheels at a 5° angle when retracted, outboard edge lower than inboard edge)
τ = τ1 + τ2 = (-2) + (-5) = -7°

Another example: 1° toe-in when lowered and leave the wheels with 2° when retracted. Inboard edge of the wheels lower than outboard edge.
τ1 = -1° (toe-in)
τ2 = 2° (leave the wheels at a 2° angle when retracted, inboard edge lower than outboard edge)
τ = τ1 + τ2 = (-1) + 2 = 1°

Finally, if you look at the first attached picture you can see an example where I set a quite extreme forward rake (ρ = 30°) and an even more extreme toe-out angle of 10° with the wheel sitting horizontal when retracted (τ1 = 10°, τ2 = 0, so τ = 10°). As you can see, everything works perfectly. The second picture I used more realistic parameters: rake = 5° (ρ = 5°), toe-in 2°, wheel sitting in its well at a 5° angle with the outboard edge lower than the inboard edge (τ1 = -2°, τ2 = -5°, so τ = -7°). Again, everything works.

So, to sum it up:

ρ: the angle of the gear leg when lowered. Positive values mean that the leg points forward, negative values mean that the leg points backward. A value of 0° means that the leg is perfectly vertical
τ: over-rotation parameter (used to account for toe-in / toe-out and/or to leave the wheels in their Wells at an angle. See text).
α = (90 - ρ) / 2
β = atan(sin(α) * tan((90° - τ) / 2))

There it is. Theoretically one could also account for camber and other parameters of the landing gear, but this is getting very complex and my brain is smoking enough
With some creativity it should be possibile to use these formulas for landing gears where the leg is not horizontal when retracted and maybe for other situations as well.

I hope that this writeup helps someone someday avoiding the usual trial and error approach for these situations.

Radial power

The alternative way

One more thing: I realized that using α and β directly may not be the easiest way to build such a landing gear. It may be preferable to use some other angles that are more practical (easier to check).
So take a look at the attached picture. Instead of using α and β to locate the oblique rotation axis, one can use γ and θ.

γ can be computed using the formula in the previous post:
γ = (90° - τ) / 2

θ can be computed as follows:
θ = atan(AC / OA)
= atan((AB / cos(γ)) / L)
= atan((L * tan(α) / cos(γ)) / L)
= atan(tan(α) / cos(γ))

The end result is the same as one would get using α and β, it's just a different way to get to the same result. Use the one you prefer.

BobH

My Feedback: (2)

Think F4U corsair. or P-40. Similar wheel movements on both of those planes.

Radial power

I'm not 100% sure, but I don't think it's the same. With the F4U and P40 you have the gear leg retracting "straight up", rotating around an axis that is perpendicular to the leg itself (no side movement), plus some additional mechanisms that rotate the wheel around the axis of the leg. The Mig-29 and Su,27, OTOH, have their gear legs move slightly outward as they retract before returning parallel to the longitudinal axis of the plane. And no additional mechanism to explicitly rotate the wheel.

That difference would be evident if someone had a video of these airplanes actuating their landing gear seen from straight ahead (or straight behind). I tried to find one such video for the Mig-29 and Su.27, but the ones in the first post are the only ones I could find.

BobH

My Feedback: (2)

Watching the video I didn't notice any side movement of the legs. Not saying it doesn't happen.
Something rotates the wheel/axel. It may be a different something that Han was used on the F4u but still it's still something lol.
The F18 Hornet also has some tricky gear geometry.

BobH

My Feedback: (2)

Watching the video I didn't notice any side movement of the legs. Not saying it doesn't happen.
Something rotates the wheel/axel. It may be a different something that Han was used on the F4u but still it's still something lol.
The F18 Hornet also has some tricky gear geometry.

Radial power

I guess that a picture is worth a thousand words, so here are 3 pics

These pics show a comparison between the movement of a Mig-29 - like gear and a Corsair-like gear. In the first two pics the Mig-29 gear is on the left and the Corsair gear on the right. The first pic shows the movement as seen longitudinally (standing in directly front or behind the airplane), while the second pic shows the same thing as seen from the bottom (from the bottom it's easier to see the position of the leg. From the top it would be mostly covered by the wheel). Please notice that the Corsair gear is reversed when compared to the Mig gear: in the Mig-29 (and Su.27) the main gear retracts forward, while on the Corsair it retracts backward. I rotated the two system so that they are side by side for a better comparison.

I hope these pics make it clear now what I mean by "side movement": If you look at the Corsair gear, you can see that its gear leg always appear vertical. With the Mig gear, instead, it visibly arcs out as it retracts.

The last pic is intended as a "proof" when compared to the first video, the one of the Mig-29 actuating its gear under power. Look at the lower part of the gear movement in the video: the gear leg appears to move in a plane parallel to the camera. In the upper part of its movement, OTOH, it visibly moves sideways. In the last pic I tried to get a similar angle and I used wireframe mode to clearly show the gear leg. You can see that the Mig-29 gear prodices exactly the movement seen in the video, while the Corsair gear looks different (notice that in this pic the positions of the 2 gear are swapped: the Corsair gear is to the left, the Mig gear to the right).

BobH

My Feedback: (2)

Yes they are different to an extent. The difference would be mainly where the top pivot point is attached. One just pivots around its axis and the other pivots and moves about its axis.
Want to see some retracting gear take a look at this lol

Radial power

Actually, that looks even simpler than the Corsair or Mig gear. Unless I'm mistaken, that gear just rotates around an axis that is horizontal and set at 45° to the aircraft longitudinal axis. The key is that the legs point straight out when extended and straight back when retracted. By the way, those legs look real thin...

Now, this is actually a complex landing gear (yes, I have a thing for Soviet/Russian gear, pun intended ):