Recent content by solvejskovlund

Thank you for this great way to explain whats happening. If I understand this correctly, the resistance of both materials will increase with increased frequency, and this increase happens individually of each other, so the copper (solid) core resistance would be the same no matter if the shield...

As I said above: So the skin effect will not cause all the electrons to prefer the aluminum path to such extent that the higher resistance of the alu will cause total resistance to increase as the copper in center is left nearly unused?

I think all of those cables can be used to send a current in one end, and you'll get it out at the other end. Sure they have different "advanced" properties - like wire cross section area, internal interference, how much current they handle, how much voltage you can apply before you may...

That assumes the shield was connected to ground, which I though was clearly stated in the question - and the follow up - that the shield was not connected to anything other than the wire (cable core) in both ends.

It's quite similar to a coax, yes. Just that the shield in a coax tend to be woven copper, while the cable I was handling, and got me into this thinking, was a aluminum sheet wrapped around. Either I don't understand what you are explaining, or we are thinking two very different scenarios. What...

Well, if the shield was connected to something else, then surely that something would be affected as well. But I was thinking of what happens to just the cable, for the simplification used as a 1-wire only. This implies that there must be another path/cable for the return current.

While I was rolling out a shielded cable, a though came to my mind - what happens to the current flow in the cable if there came a short between the wire and the shield in both ends of the cable? For simplicity, lets assume a 1-wire copper wire wrapped in an aluminum shield. The wire and the...

Sorry for the late reply - I lost my phone - only way of internet access.I don't have any access to make changes to the MC behavior. Hence everything has to be a separate logic block.I'm also thinking of dropping the MC signal. The MC makes its decision based on voltage from a sensor. It would...

I'm need to control a relay in three ways: 1: Forced OFF signal. This is a signal triggered from a MC that I have no access to change. It is a 1 wire output that is normally floating, and pulled to ground for exactly 1 minute when the MC has detected that the relay should turn off. Important...

A unknown x given by another unknown θ that is derived from x, doesn't solve anything. I'm surprised this wasn't straight forward math.

For (5) I had ## x^2 -2R\frac{L}{5}+\frac{L^2}{25}=0 ## Solving (5) for R: ## 2R\frac{L}{5}=x^2+\frac{L^2}{25} ## ## R=x^2\frac{5}{2L}+\frac{L^2}{25}*\frac{5}{2L} ## ## R=\frac{5x^2}{2L}+\frac{L}{10} ## Insert into (4) ## x=(\frac{5x^2}{2L}+\frac{L}{10})*sin(\Theta) ## ##...

Thought you ment something that wasn't already listed. (4) gives ## R=\frac{x}{sin(\Theta)} ## (5) gives ## x^2 + R^2-2R\frac{L}{5}+\frac{L^2}{25}=R^2 ## ## x^2 -2R\frac{L}{5}+\frac{L^2}{25}=0 ## combined: ## x^2 -2* \frac{x}{sin(\Theta)} *\frac{L}{5}+\frac{L^2}{25}=0 ## Having x and θ as...

Uh. I didn't realize I had left out the /2 when using θ. Glad you pointed that out. I don't see which trig relationship you're pointing to.

R-h=R*cos(θ) ## cos(θ)=\frac{R}{R}-\frac{h}{R}=1-\frac{L}{5R} ## Could also be written as ## \frac{L}{5R}=1-cos(θ) ## ## \frac{L}{R}=5-5*cos(θ) ## From before I have: ## \frac{L}{2πR} = \frac{2θ}{2π} = \frac{θ}{π} ## (segment length vs circle length = angle vs circle angle) ## \frac{L}{R} =...

Some formulas I see related: ## \frac{L}{2πR} = \frac{2θ}{2π} = \frac{θ}{π} ## (segment length vs circle length = angle vs circle angle) R2 = x2 + (R-L/5)2 (pythagoras) ## x=R*\sin(\frac{\Theta}{2}) ## Combining these, knowing that x=2c, and h=L/5, I just get the formulas of the wiki as...