# The Physics Behind the American Death Triangle [closed]

I've heard a lot about the American Death Triangle and how it is awful for belaying. The Death Triangle is set up as such: you have two anchor points with a single rope or line running through both points. The rope then connects to a single harness at both ends.

The anchors don't act like a single fastening point, but instead a bit like pulleys in that they redirect the force. The lines do not stop on the anchors. As such, while the force is evenly distributed between the anchors, it is magnified because there is the force of the weight as well as the force of the tension between the anchors. The equation for each anchor is given by

$$F = \frac{W}{\cos\left(45 ^\circ + \frac{1}{4} \theta\right)}$$

Where $\theta$ is the angle of the bottom of the triangle. The equation came from Wikipedia, but what I don't understand is how they derived this equation. It has no sources nor explanation nor proof. Can walk me through the proof of this equation?

## closed as off-topic by CuriousOne, ACuriousMind♦, Kyle Kanos, Bill N, John RennieMar 23 '16 at 6:34

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This is a great example of elementary physics concepts, which perhaps due to my lack of skill bring in some rather inelegant mathematics. The whole concept can be explained with a free body diagram.

First, let's derive the first expression they give on the wikipedia page for a "V" formation. This is perhaps a classic freshman physics problem with free diagrams; because it is simple enough to do, but challenging enough for a freshman not to simply guess the answer. Below I have drawn the two situations on the same graphic. For the "V" formation, consider there is only rope where the blue lines are. Then for the death triangle, add rope across the orange line. The situation as described is an isosceles triangle, with two interior angles $\phi$ and $\theta$. The circles are the anchors (assumed massless), and the square is the load of mass $m$.

Now write out all the forces for each case, and solve the system of equations.

# "V" formation

On the anchor there are two forces: i) from the tension in the rope due to the object (pointing along the blue line), and ii) the rock it is wedged in providing the necessary force to hold the anchor in place. We wish to solve for the force the rock needs to apply to the anchor to hold it in place, i.e., the sum of the two forces is 0 (thus the force from the rock, $F_\textrm{rock}$, equivalent to $F_{Anchor}$ in wikipedia notation, points equal and opposite to the tension). We therefore just need to solve for the tension, $T$, which is most easily done from the free body diagram of the load.

The load has three forces acting on it:

• gravity,

• the tension from the left anchor,

• the tension from the right anchor.

Now this is a 2 dimensional problem, so let's break down the force equations by components. Looking at the $x$ component

$$\left( \sum_{i} \vec{F}_i \right)_x = -T_{\textrm{left}} \sin\left(\frac{\theta}{2}\right) + T_{\textrm{right}} \sin\left(\frac{\theta}{2}\right) = 0.$$

Therefore we see that $T_{\textrm{left}}=T_{\textrm{right}}$, which I will now just call $T$. This is what you would expect from the symmetry of the problem. Now the $y$ component says,

$$\left( \sum_{i} \vec{F}_i \right)_y = -mg + T \cos\left(\frac{\theta}{2}\right) + T \cos\left(\frac{\theta}{2}\right) = 0.$$

Thus we have arrived at $$T = \frac{mg}{2 \cos\left(\frac{\theta}{2}\right)}.$$

As already stated the magnitude of the force the rock exerts, is equal to the tension; therefore, this tension is the our answer for $F_\textrm{rock}$.

Note to make the math simpler in the next section that $\cos(x) = \sin(\pi/2 - x)$, and that $2\phi = \pi - \theta$, therefore $\cos(\theta/2) = \sin(\phi)$ (this is obvious if you split the triangle into two right triangles by bisecting $\theta$).

# Death Triangle

Now we include the orange bit of rope, such that it is all connected together. Similar to the "V" formation, we wish to solve for the force that the rock exerts on the anchor to keep it in place. Again this force will be equal and opposite of all other force, such that the net force sums to zero. However, this time the anchor feels three forces:

• tension from the rope attached to the load,

• tension from the rope attached to the other anchor,

• the keeping force of the rock. Note this time that the keeping force of the rock is not along the triangle, since it must balance out both (i) and (ii).

Since the anchors act like pulleys you know that they just redirect the force, without modification. Thus the tension the anchor feels from the other anchor is equal to the tension it feels from the object. Therefore we will call all tension forces simply $T$. An intuitive way of seeing this is a counter-example. Consider the two forces were not equal, then the side of the rope with the bigger force would pull the other rope towards it. We know that these ropes are at rest, thus we intuitively know these forces must be equal. With this in mind we write the $x$ component of the force equations as

$$\left( \sum_{i} \vec{F}_i \right)_x = T + T \cos\left(\phi\right) - F_x = 0.$$

Where $F_x = F_\textrm{rock} \cos(\phi')$, where $\phi'$ is the angle $\vec{F}_{rock}$ makes with the $x$-axis. I will not worry too much about it since we will use both equations to replace it with known variables. Therefore we move onto the $y$-component

$$\left( \sum_{i} \vec{F}_i \right)_y = T \sin\left(\phi\right) - F_y = 0.$$

Where again similarly $F_y = F_\textrm{rock} \sin(\phi')$. Thus setting these equations equal to each other and solving for $\phi'$ in terms of $\phi$

$$\phi' = \arctan\left( \frac{\sin(\phi)}{1+\cos(\phi)}\right).$$

Now we can solve for $F_{rock}$ by rewriting the $y$-equation as

$$F_\textrm{rock} = T \frac{\sin(\phi)}{\sin(\phi')}.$$

Now we know $T$ from doing the "V" formation, and recall at the end we noted how to rewrite it in terms of $\sin(\phi)$, which now cancel out giving us

$$F_\textrm{rock} = \frac{mg}{2\sin(\phi')}.$$

This is the exact same form of the equation as the "V" formation, just with $\phi$ and $\phi'$ switched! Now $\sin\left( \arctan\left( \frac{\sin(\phi)}{1+\cos(\phi)}\right) \right)$, is a slightly unsightly sight. But if you only consider, $0 \leq \theta \leq \pi$, as your range for $\theta$, which are limits of a triangle, then it simplifies to to $\sin\left(\frac{\phi}{2}\right)$. Now recalling that $2\phi = \pi - \theta$, then we can rewrite that as $\cos\left(\frac{\pi}{4} + \frac{\theta}{4}\right)$. Thus we have indeed arrived at

$$F_\textrm{rock} = \frac{mg}{2\cos\left(\frac{\pi}{4} + \frac{\theta}{4}\right)}.$$

I haven't spent much time trying to make the math elegant, but you can see the agreement, which leads us to believe we have included all the physics in which the wikipedia article has.

# Conclusion

As noted in the wikipedia, this formation is less effective at minimizing $F_\textrm{rock}$. The "V" formation, best case scenario of $\theta = 0$, halves weight; while for the death triangle you only get $\sqrt{2}/2$. Therefore, for safety reasons, where you want to put the smallest amount of weight on an anchor, it is advised to not use this method of anchoring (on top of apparently violating other tenants of climbing, e.g., redundancy).

• Just a small remark on the "V formation". It is plausible to assume that $T_\mathrm{left}=T_\mathrm{right}$ since your lower point has no friction. What your $x$ equation there actually proves is that $\theta=\theta'$: your forces are in an equilateral triangle. – Nick Jun 9 '16 at 6:56