Why doesn't the Falcon-9 first stage use three legs to land?̀l D

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enter image description here The immediate thought that would probably come into your mind would be "Because 4 legs is more stable than 3." However that is not always true. 3 legs offer the same or in some cases more stability as 4 legs as 3 legs guarantee they are in the same plane. This Stack Exchange answer to the question Why did the Apollo Lunar Module have four landing legs? states that,

Three legs offers less safety margin for steep slopes and sideways landings.

However the boosters always land on a flat smooth surface and even if the booster did land with a lot of lateral velocity, it would be damaged or fall into the ocean. Also with three legs, the booster would be even lighter and cheaper.

Question: So why did SpaceX choose to use 4 legs on a Falcon 9 booster? Was it simply because it offers more stability in some cases (such as landing sideways) and has less margin for error, or does it simply distribute more weight, or is it another, less obvious reason?

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I can't speak for why SpaceX made the decision. However, while three legs won't wobble, four legs are less likely to tip over. SpaceX has demonstrated tipping over is a major problem.

Dr Peterson of The Math Forum explains...

There are different kinds of stability! A three-legged stool is guaranteed not to wobble, because the ends of its legs always form a plane. But a little wobble is only an inconvenience. More important for practical purposes, [a three legged] stool is LESS stable than one with more legs in the sense that its center of gravity is further inside its base: the more sides a regular polygon has, the greater its apothem (the distance from the center to the middle of an edge). That greater distance means that the sitter can lean farther out in any direction without tipping over. So if you don't mind a little tipping but don't want to fall on your face, or if you have a reasonably even floor, more legs are better.

To do the math, imagine the circle made by the rocket legs is of radius 1 for simplicity. Three landing pads are on the vertices of an equilateral triangle. Four landing pads are on the vertices of a square.

enter image description here

Let's all pretend that's an equilateral triangle. The blue lines are the apothems. The green line is the radius. They form a triangle with an inner angle of 120/2 or 60 degrees. We can solve for the apothem using the law of sines.

$$\\frac{a}{\\sin 30} = \\frac{1}{\\sin 90}$$ $$a = \\frac{1 * \\sin 30}{\\sin 90}$$ $$a = 0.5$$

And now four legs.

enter image description here

Same idea, but now the angle is 45 degrees.

$$\\frac{a}{\\sin 45} = \\frac{1}{\\sin 90}$$ $$a = \\frac{1 * \\sin 45}{\\sin 90}$$ $$a = 0.707$$

At five legs the apothem is 0.809. At six it's 0.866. The basic formula is $\\sin(90 - \\frac{180}{n})$.


But what if we just used three longer legs? Would that be less weight than four shorter ones? In other words, we need to get the three legged rocket's apothem to 0.707. How much further spread out do the landing pads have to be? Set a to 0.707 and solve for r.

$$\\frac{r}{\\sin 90} = \\frac{0.707}{\\sin 30}$$ $$r = \\frac{0.707 * \\sin 90}{\\sin 30}$$ $$r = 1.414$$

All three landing pads need to be over 40% more spread out than four. For three legs that's 120% further, and that's before we consider that the legs are at an angle and so need to be considerably longer to have the pads go 40% further out from the rocket's center. Being longer they would need to be stronger and even heavier.

More legs provide rapidly diminishing returns. The general formula is simply $\\frac{a2}{a1}$ or $\\frac{\\sin angle2}{\\sin angle1}$.

  • For 3 legs to match 4. $\\frac{\\sin 45}{\\sin 30}$ or 1.414.
  • For 4 legs to match 5. $\\frac{\\sin 54}{\\sin 45}$ or 1.144
  • For 5 legs to match 6. $\\frac{\\sin 60}{\\sin 54}$ or 1.070

Four legs only need to be 15% broader to match the stability of five, or 60% total, making it likely more economical to use four longer, stronger legs than five shorter ones.

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    $\\begingroup$ This is a good point especially if you consider it has to land on a very small boat. In order for three legs to have the same resistance to tipping, they would have to extend farther away from the rocket and so increase the chance of missing the edge of the boat, not to mention they'd have to be longer and/or stronger+heavier depending on the implementation details. $\\endgroup$ – uhoh 8 hours ago
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    $\\begingroup$ @uhoh Good point. I'll work out the math for how much longer three legs have to be to give equivalent stability as four. $\\endgroup$ – Schwern 7 hours ago
  • $\\begingroup$ Looks beautiful! $\\endgroup$ – uhoh 7 hours ago
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    $\\begingroup$ @uhoh Done. Four is the magic number. $\\endgroup$ – Schwern 7 hours ago
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A stool with three legs that are rigid, are of a fixed length, and have a fixed orientation with respect to the seat of the stool is superior to a chair with more than three rigid, fixed length, fixed orientation legs in one and only one regard. And that one regard is completely irrelevant in the case of landing legs.

The landing legs on any landing space vehicle inevitably violate at least one of those three conditions (rigidity, fixed length, fixed orientation) that are needed to make a three legged stool superior.When a Falcon first stage does land, all of the legs will be on the surface due to these features. While three legs buys very little with respect to stability, a three legged lander loses a whole lot with respect to tippability.

Another factor is that the Falcon has nine engines, one in the center and the other eight around it in an octagonal or square-like arrangement. The bottom part of the thruster reflects this tetradic / octagonal symmetry. The bottom attachment points are close to the engines and have to fit within the constraints of the engine layout. Having four landing legs obeys this symmetry; having only three would not.

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    $\\begingroup$ "Another factor is that the Falcon has nine engine..." for just a split second, I thought you were going for a Feng shui-based argument, but I should have known better. $\\endgroup$ – uhoh 3 hours ago

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