Can RSE Line(tm) Support Project TouCans as a Dipole

 I've been following the introduction of RSE wire on VE6LK's YourTube channel. It's pretty spiffy stuff. It has nickel coated carbon fibers as its core and a rope-like insulator. Basically, it's cable that behaves like a rope. It can also support up to 300 pounds without breaking.

The tensile sttrength is what caught my attention. Could RSE be used as a dipole with Project TouCans? Remember, when TouCans is used with a dipole, all 22 ounces or so of the rig go up in the dipole as well.

I haven't worked out how to actually connect the wire to the antenna terminals yet, but assuming I could, and I pulled the RSE taught over a couple of tree branches, could it hold up the rig?

It's time to do some physics.

I asked GPT-5.1 to do the physics for me. Here's my prompt:

I'd like to do a physics problem with you. I'm doing this to find out  if we can hold a radio in the middle of a dipole. The wire has a breaking ension of 300 pounds. The radio weighs 2 pounds. Obviously, the wire can pick up the radio when the wire is straight  above the radio. Here's what we need to figure out/demonstrate:

1. Draw a free force diagram with the radio as a 2 pound mass with the acceleration of gravity in the middle. Two lines of wire project outwards from the radio. The wire is symetric, so both strands will be under tension T.

1.a. Lable the angle between the wire and the horizontal as theta (in greek letters)

2. Derive the Tension T in terms of theta and the mass of the radio m. Show all steps.

3. Use the derived formula to determine the minimum anlge theta before the tension T would break the wire which breaks when suspending a 300 pound weight. 

Here's the answer I got back:

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Nice problem 😄 This is exactly the classic “why a nearly horizontal wire explodes” setup.

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1. Free-body diagram (FBD)

We’ll treat the radio as a point mass hanging at the middle.

Forces on the radio:

• Weight: $\vec{W} = m\vec{g}$, straight down (magnitude $W = mg$, or 2 lb in your numbers).

• Tension from left wire: magnitude $T$, along the left wire, up and out.

• Tension from right wire: magnitude $T$, along the right wire, up and out.

You defined:

• $\theta$ = angle between each wire and the horizontal.

So each tension vector $\vec{T}$ makes an angle $\theta$ above the horizontal.

In text form, the FBD looks like:

• Downward arrow from the radio: $W = mg$

• Two arrows symmetric about the vertical:

  • Left: $T$ at angle $\theta$ above horizontal to the left

  • Right: $T$ at angle $\theta$ above horizontal to the right

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2. Derive $T$ in terms of $\theta$ and $m$

We’ll do equilibrium in horizontal ($x$) and vertical ($y$) directions.

Components of the tensions

Because $\theta$ is measured from the horizontal:

• Horizontal component of each tension: $T\cos\theta$

• Vertical component of each tension: $T\sin\theta$

Directions:

• Left wire:

  • $T_x = -T\cos\theta$

  • $T_y = +T\sin\theta$

• Right wire:

  • $T_x = +T\cos\theta$

  • $T_y = +T\sin\theta$

Weight:

• $W_x = 0$

• $W_y = -mg$

Equilibrium equations

Because the radio is at rest:

$$
\sum F_x = 0,\quad \sum F_y = 0
$$

Horizontal:

$$
\sum F_x = -T\cos\theta + T\cos\theta = 0
$$

(as expected, they cancel)

Vertical:

$$
\sum F_y = T\sin\theta + T\sin\theta - mg = 0
$$

$$
2T\sin\theta = mg
$$

Solve for $T$:

$$
T = \frac{mg}{2\sin\theta}
$$

If you prefer to work directly in pounds as forces:

• Let $W$ = weight of the radio = 2 lb.

• Then:

$$
2T\sin\theta = W \quad\Rightarrow\quad
T = \frac{W}{2\sin\theta}
$$

---

3. Minimum angle $\theta$ before the wire breaks

Each strand of wire breaks when the tension reaches 300 lb. So we require:

$$
T \le 300\ \text{lb}
$$

Using the formula $T = \dfrac{W}{2\sin\theta}$ with $W = 2\ \text{lb}$:

$$
\frac{2}{2\sin\theta} \le 300
$$

$$
\frac{1}{\sin\theta} \le 300
$$

$$
\sin\theta \ge \frac{1}{300}
$$

So the minimum angle $\theta_{\min}$ is:

$$
\theta_{\min} = \arcsin\left(\frac{1}{300}\right)
$$

Numerically:

• $\dfrac{1}{300} \approx 0.003333$

• For small angles, $\sin\theta \approx \theta$ in radians, so:

$$
\theta \approx 0.003333\ \text{rad}
$$

Convert to degrees:

$$
\theta_{\min} \approx 0.003333 \times \frac{180}{\pi}
\approx 0.19^\circ
$$

So:

The wire can safely hold the 2-lb radio as long as each side makes at least about $0.19^\circ$ above the horizontal.

Equivalently: if the wire is even slightly above perfectly horizontal, the tension is far below the 300 lb breaking strength. As $\theta \to 0^\circ$, the tension $T = \dfrac{W}{2\sin\theta} \to \infty$, which is the usual “danger” with very shallow angles.

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So! RSE can hold TouCans. I've never come close to tugging on the existing wire hard enough to make the dipole have less of an angle with the horizontal of less than 10 degrees. This should be fine.

Next, I need to figure out how to connect RSE to the rig itself.

References

Food for thought: Soldering RSE

Thoughts on Using LLMs (AI) for Physics

LLMs can be a great help with both physics problems IRL like this one, and with physics homework. As with all LLM output, you need to check it. The derviation above is correct. I know because I've taken classes up through graduate level mechanics.

Having said that, there are several advantages to using LLMs for this sort of thing. The LLM made this post as easy as dashing off a set of calculations on the back of an envelope. Without the LLM this post probably woulnd't exist. The LLM also gave me the space to think through the solution. I've never had that kind of space before since I was always eith in lectures, or the stakes were higher becuase I was doing homework for a grade. Thanks to getting that space, I now understand tension calculations better than I did when I started this post. I've internalized what they mean, whereas before I was simply running a rote process. (It was always the x forces canceling that really bothered me, but I can now see that the equation setup for the y components contains the magnitude of the tension, (much of which is in the x direction), and therefore while the x components 'canceled out' there is in fact a lot of tension in the x direcation where it intuitively feels like it should be.)