Home |
Search |
Today's Posts |
|
#1
![]() |
|||
|
|||
![]()
You're correct, I neglected proximity effect. And the wires are close
enough that it's a factor. What is the approximate expression you used? Roy Lewallen, W7EL Reg Edwards wrote: As he says, Roy's resulta are very approximate. That's mainly because he neglected proximity effect between the wires. A better approximation is obtained by incorporating an approximate expression for proximity effect. However, this makes differentiation of the loss formula with respect to wire diameter ridiculously tedious. So I found minimum loss by plotting a graph with a pocket calculator and searching for it. . . . |
#2
![]() |
|||
|
|||
![]() Provided skin effect is fully operative, ie., skin depth is about 1/6th wire diameter or less, proximity effect increases wire resistance by dividing normal skin-effect resistance of a single straight wire by K : K = SquareRoot( 1 - Square( D / S ) ) where D is wire diameter and S is centre-to-centre wire spacing. Note that resistance increases towards infinity as the pair of wires approach contact with each other. This is confirmed by precision measurements. To minimise line attenuation for any given wire spacing, maximise U with respect to D : U = D * InvCosh( S / D ) * SquareRoot( 1 - Square( D / S ) ) . ----- Reg, G4FGQ |
#3
![]() |
|||
|
|||
![]() Provided skin effect is fully operative, ie., skin depth is about 1/6th wire diameter or less, proximity effect increases wire resistance by dividing normal skin-effect resistance of a single straight wire by K : K = SquareRoot( 1 - Square( D / S ) ) where D is wire diameter and S is centre-to-centre wire spacing. Note that resistance increases towards infinity as the pair of wires approach contact with each other. This is confirmed by precision measurements. To minimise line attenuation for any given wire spacing, maximise U with respect to D : U = D * InvCosh( S / D ) * SquareRoot( 1 - Square( D / S ) ) . ==================================== Roy, having given a little more thought to it, I think that by differentiating U with respect to D and equating dU/dD to zero, things will then simplify and the value of the ratio S /D, and hence Zo, will be obtained directly. If you still have enough enthusiasm I leave it to you to perform the differentiation. ---- Reg. |
Reply |
Thread Tools | Search this Thread |
Display Modes | |
|
|
![]() |
||||
Thread | Forum | |||
Phone line as SW antenna [04-Apr-00] | Info | |||
Ladder line question ???? | Antenna | |||
Folded monopole dilemma | Antenna | |||
Phone line as SW antenna [04-Apr-00] | Info | |||
Current in antenna loading coils controversy - new measurement | Antenna |