I'm building a swing set for my children. All of the designs I've seen involve building two A-frames and connecting them at the top with a crossbar/beam from which hang the swings. The A-frames are placed perpendicular to the ground. In addition, it seems that the A-frames are commonly constructed at an angle of 12 degrees from the perpendicular to the ground.

I have several questions:

  1. Where does the 12 degree angle come from?

  2. I've been on swingsets that move too much. I think this is due to a combination of unanchored posts and the angle between the legs of the frame being too little. How do I compute the angle that will provide optimum stability?

  3. Shouldn't the frames be placed at an angle with respect to the ground (e.g. 70 degrees) rather than perpendicular, in order to prevent side-to-side motion?

  4. How would I go about computing the above?

Assumptions that I'm going on include:

  • 2 swings
  • Adults will swing (up to 200 lbs)
  • A-frames will be anchored to concrete piers
  • 1
    $\begingroup$ The out-of-the-swinging-plane bit is addressed with one of (1) Some are on the angle, (2) some have a angled third support, and (3) the fact the the supports are typically pipes of non-trivial diameter allows a considerable amount of stress to be taken up in the joints (this kind of thing is a common analysis in a Engineering Statics class). $\endgroup$ Jul 23, 2012 at 21:27

3 Answers 3


Using a simple pendulum approximation, at the point of , the maximum tipping force occurs when the swing is at $\theta=45°$ angle. In general the forces on the bar are

$$ A_x = -\cos\theta \left( m g \sin \theta + \frac{\tau}{L} \right)-m L \dot{\theta}^2 \sin \theta $$ $$ A_y = m \cos \theta \left( g \cos \theta + L \dot{\theta}^2 \right) - \frac{\tau}{L} \sin \theta $$

where $L$ is the chain length, $\tau$ is the torque applied by your back on the pelvis (pull up torque) and $m$ is the mass of the swinger.

If $h$ is the height of the bar from the ground, and $b$ the base distance across the A-frame then the forces on supports are

$$ G_x = -A_x $$ $$ G_y = \frac{1}{2} A_y - \frac{h}{b} A_x $$ $$ H_y = \frac{h}{b} A_x + \frac{1}{2} A_y $$

Combining you will see that the forces in the frame are not always compressive, but tensile also. In the above expression $H_y$ is always negative (frame pulling on the ground) when $\theta > (\text{frame angle})$.

So as far as physics goes, there is no significance to the 12°. Engineering wise, I like the other answer of minimizing the material used.

  • $\begingroup$ Could you clarify Gx, Gy, and Hy? $\endgroup$
    – Hendy
    Apr 26, 2020 at 18:55
  • 1
    $\begingroup$ If A is the pivot point of the swing, then $A_x$ and $A_y$ are the connection forces along with the two directions. Similarly, for the two support points G and H which are in front and on the back of the swing center. A diagram would have helped I guess. $G_x$, $G_y$ and $H_y$ are support forces from the ground. $\endgroup$ Apr 26, 2020 at 20:00

The greater the angle in the A-frame, the more metal you'll have to use. This, balanced with wanting to keep the swing set firmly planted probably led to the heuristic 12 degrees.

(EDIT: ignore gravity transmitted through the chain because it doesn't ever apply torque to swing set.) If you want to calculate some forces on the swing, the force a swinger exerts on the top bar is along the swing chains and equal to mass*speed^2/length of chain. The speed is minimal when the swing chains are parallel to the ground, and this is also when the direction of the swinger's force is delivering the maximum torque to tip over the swing set. From this argument, I think the tipping force is not much.

  • $\begingroup$ This is wrong, here is simple proof. A swinger just sitting in swing has speed=0. According to this answer, his force is zero but we know his force is $mg$ where m is his mass and g is the acceleration due to gravity. $\endgroup$
    – mwengler
    Jul 24, 2012 at 15:23
  • $\begingroup$ I do not care about the force of gravity on the swing set transferred through the chain because it does not contribute any torque to tipping over the swing set. $\endgroup$ Jul 24, 2012 at 18:23
  • $\begingroup$ OK, on ignore gravity. But your statement is still wrong. The torque is generally maximum at some point before the chain is parallel to the ground. If swing even gets high enough for chain to be parallel, usually it is with close to 0 speed at that point since this is maximum height of swing. Therefore, maximum torque is some point BEFORE maximum height, which is virtually always before chain is parallel to ground. $\endgroup$
    – mwengler
    Jul 24, 2012 at 18:46
  • $\begingroup$ I know this. Please read my answer again. $\endgroup$ Jul 25, 2012 at 2:45

Thinking of it as a pendulum is the best answer but I don't agree that the major tipping force will be at 45 degrees ($\frac{\pi}4$ radians). Whilst it is clear the greastest force is excerted at 45 degrees ($\frac\pi4$ radians) in simplistic terms on the Y component is the tipping force. Thus you can make X any number you want and the swing wont tip it will simply start to crush the A frame. Therefore the greatest tipping force will be when the load is parrallel to the ground.

Now back to your question. Efficient material use is the answer. As a guide with posts in concrete the structure should be stable when you just use the weight of the components. Thus you will get extra help by the way the concrete adheree to the ground but this is difficult to calculate so should be taken as a bonus rather than a critical component. So work out the total weight of the various components and calculate the x and y components of the vector.

Note if you are burying the poles deep in the earth then you can also include a bit of lever action of the mud on the bottom of the pole but again this is unreliable and generally not constant.

Now work out the x and y components of the weight of the load and for now throw away ( place carefully on the side) the x component of both and balance up the y with at least a $2:1$ ratio. Doing it like this you can easily see the weight of the concrete and poles become silly big with an upright pole.

Using practical sizes of timber for the A frame and concrete it starts looking feasible anything above about 10 degrees. However if I were doing it for the kids ( as I have done) i'd make it at least double and probably more like 30 degrees.


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