Mathcad_-_CONDIST

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Client: Computed by:

Project: Date:

Project No.: File No.: Checked by:

Title: Calculation of Distance Between Conductors Date:

==========================================================================

These calculations determine the distance between two conductors at any point on each

conductor. By varying the point locations, defined as the distance from one support of each

conductor, the minumum distance can be found. Strain insulator droop (sag from horizontal) is

considered separately from conductor sag to provide greater accuracy, particularly for slack

spans. Distance from one conductor to a guy wire or other straight line structure, e.g. a pole, canbe found by treating the guy or structure as a second conductor with a very high tension and low

unit weight.

Catenary equations are based on equations given in "Sag-Tension Computations and Field

Measurements of Bonneville Power Administration", Paul F. Winkelman, AIEE Paper 59-900,

February 1960.

Calculations are based on a coordinate system where the left strain insulator attachment point of

conductor 1 is at x =0, y =0, z =height and the right attachment point is at x =A1 (span), y =0,

z =heght. The coordinate system, span length, and attachment heights are adjusted to account

for the effect of strain insulators. The following figure shows the definition of the coordinate

systems and some key distances. Conductor positions are first calculated on two separate

coordinate systems: x1,y1,z1 and x2,y2,z2. These coordinate systems are defined by the endsof the conductors. The left end of conductor 1 is at x1 =0, y1 =0, z1 =height. After calculating

the conductor positions in terms of x1,y1,z1 and x2,y2,z2, the coordinates are translated and

rotated to convert them to the x,y,z system.

Wind is assumed to be normal to conductor 1 (direction y1). The wind force on conductor 2 is

reduced by the cosine of the angle between the conductors. No in-line movement of conductor 2

by the wind is considered.

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Strain insulators are assumed to be rigid and of uniform weight per unit length. The wind force on

the insulators is assumed to be the same as on a solid cylinder of the same diameter. Any swing

of suspension insulators must be accounted for separately in the coordinates for the attachment

points; this is not considered in these calculations. Strain insulator length, droop, pull out, and

sideways swing (as shown on the figure below) change the effective conductor span from between

the conductor supports to between the ends of the insulators. If there are no strain insulators on

one end of a conductor, enter a value of zero for the corresponding strain insulator length. This is

also the case where the conductor is suspended from the structure on suspension insulators - the

support locations are at the connections of the conductors to the insulators.

DERIVED UNITS:

psfpsi

144

INPUT DATA:

Wind pressure

Conductor 1:

Pw 4 psf :=

Span length A1 250 ft:= Conductor unit weight v1 1.019lbf

ft:=

Conductor diameter Dc1 0.99 in:= Horizontal tension H1 1000 lbf:=

Left support height Z1L 50 ft:= Right support height Z1R 55 ft:=

Strain insulators:

weight

diameterVs1 84 lbf := length (left str)

length (right str)Ls1L 4.35 ft:=

Ds1 10 in:= Ls1R 4.35 ft:=

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Conductor 2:

Conductor unit weight v2 0.875lbf

ft:=

Conductor diameter Dc2 0.99 in:= Horizontal tension H2 3000 lbf:=

Left support height Z2L 40 ft:= Right support height Z2R 74 ft:=

Strain insulators:

weight

diameterVs2 84 lbf := length (left str)

length (right str)Ls2L 4.35 ft:=

Ds2 10 in:= Ls2R 4.35 ft:=

X,Y coordinates of supports in relation to the corresponding support of conductor 1:

left support x

left support yL2X 50 ft:= right support x

right support yR2X 50 ft:=

L2Y 50 ft:= R2Y 250 ft:=

VARIABLES:

Distance from left structure to measurement point:

Conductor 1 xc1 70.7 ft:=

Conductor 2 xc2 52.85 ft:=

CALCULATIONS:

Conductor 1:

Wind on conductor w1 Pw Dc1:= w1 0.33lbf

ft=

Wind on left insulator Ws1L Pw Ds1 Ls1L:= Ws1L 14.5lbf=

Wind on right insulator Ws1R Pw Ds1 Ls1R:= Ws1R 14.5lbf=

Unadjusted support elevation difference: Bu1 Z1L Z1R := Bu1 5 ft=

Next calculate the strain insulator positions: droop down, pull out, and swing sideways. The

conductor tension, wind load and weight used are unadjusted for the insulator droop and swing.

Distance to low point:

Au1 A1 Ls1L Ls1R:= Au1 241.3 ft=

X0u1Au1

2

H1

v1

asinh

Bu1

2

H1

v1sinh

Au12

H1

v1

+:= X0u1 100.368 ft=

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Left insulator:

droop angle v1L atan

v1 X0u1Vs1

2+

H1

:= v1L 8.21deg=

droop SZ1L Ls1L sin v1L( ):= SZ1L 0.621 ft=

swing angle t1L atan

w1Au1

2

Ws1L

2+

H1

:= t1L 2.695 deg=

swing SY1L Ls1L sin t1L( ):= SY1L 0.205 ft=

pull SX1L Ls1L cos v1L( ) cos t1L( ):= SX1L 4.301 ft=

Right insulator:

droop angle v1R atan

v1 Au1 X0u1( )Vs1

2+

H1

:= v1R 10.515 deg=

droop SZ1R Ls1R sin v1R( ):= SZ1R 0.794 ft=

swing angle t1R atan

w1Au1

2

Ws1R

2

+

H1

:= t1R 2.695 deg=

swing SY1R Ls1R sin t1R( ):= SY1R 0.205 ft=

pulll SX1R Ls1R cos v1R( ) cos t1R( ):= SX1R 4.272 ft=

Now make adjustments for insulator droop, pull and swing.

Span length Aa1 A1 SX1L SX1R:= Aa1 241.427 ft=

Left support height Za1L Z1L SZ1L:= Za1L 49.379ft=

Right support height Za1R Z1R SZ1R := Za1R 54.206ft=

Support height difference B1 Za1L Za1R := B1 4.827 ft=

Slope distance C1 Aa12

B12

+:= C1 241.475 ft=

Distance to low point X01Aa1

2

H1

v1asinh

B1

2

H1

v1

sinh

Aa1

2

H1v1

+:= X01 101.142 ft=

Height at low point Z01 Za1LH1

v1cosh

X01

H1

v1

1

:= Z01 44.162ft=

Distance to wide point Xw1Aa1

2:= Xw1 120.714 ft=

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Maximum conductor blowout Yw1H1

w1cosh

Xw1

H1

w1

1

:= Yw1 2.405 ft=(not including insulators)

Height at measurement point Z1 x1( ) Z01H1

v1cosh

x1 X01

H1

v1

1

+:=

Blowout at measurement point

Y1 x1( ) Yw1H1

w1cosh

x1 Xw1( )C1

Aa1

H1

w1

1

SY1L+ SY1L SY1R( )x1

Aa1:=

X-coordinate X1 x1( ) x1 SX1L+:=

Position vector cond1 x1( )

X1 x1( )

Y1 x1( )

Z1 x1( )

:= cond1 xc1( )

75.001

2.196

44.634

ft=

Graph the conductor location: xc 0 ft Aa1..:= Aa1

40 20 0 20 4046

48

50

52

54

Z1 xc( )

xc

SAG

40 20 0 20 402

1

0

1

2

Y1 xc( )

xc

BLOWOUT

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Conductor 2:

Span length A2 A1 R2X+ L2X( )2

R2Y L2Y( )2

+:= A2 335.41 ft=

Angle from conductor 1 12 atanR2Y L2Y

A1 R2X+ L2X

:= 12 63.435deg=

Wind on conductor w2 Pw Dc2 cos 12( ):= w2 0.148lbf

ft=

Wind on left insulator Ws2L Pw Ds2 Ls2L:= Ws2L 14.5lbf=

Wind on right insulator Ws2R Pw Ds2 Ls2R:= Ws2R 14.5lbf=

Unadjusted support elevation difference: Bu2 Z2L Z2R := Bu2 34 ft=

Next calculate the strain insulator positions: droop down, pull out, and swing sideways. The

conductor tension, wind load and weight used are unadjusted for the insulator droop and swing.

Distance to low point:

Au2 A2 Ls2L Ls2R:= Au2 326.71 ft=

X0u2 Au2

2

H2

v2asinh

Bu2

2

H2

v2sinh

Au2

2

H2

v2

+:= X0u2 192.674 ft=

Left insulator:

droop angle v2L atan

v2 X0u2Vs2

2+

H2

:= v2L 2.416 deg=

droop SZ2L Ls2L sin v2L( ):= SZ2L 0.183 ft=

swing angle t2L atan

w2Au2

2

Ws2L

2+

H2

:= t2L 0.599 deg=

swing SY2L Ls2L sin t2L( ):= SY2L 0.045 ft=

pulll SX2L Ls2L cos v2L( ) cos t2L( ):= SX2L 4.346 ft=

Right insulator:

droop angle v2R atan

v2 Au2 X0u2( )Vs2

2+

H2

:= v2R 9.397 deg=

droop SZ2R Ls2R sin v2R( ):= SZ2R 0.71 ft=

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swing angle t2R atan

w2Au2

2

Ws2R

2+

H2

:= t2R 0.599 deg=

swing SY2R Ls2R sin t2R( ):= SY2R 0.045 ft=

pulll SX2R Ls2R cos v2R( ) cos t2R( ):= SX2R 4.291 ft=

Now make adjustments for insulator droop, pull and swing.

Span length Aa2 A2 SX2L SX2R:= Aa2 326.773 ft=

Left support height Za2L Z2L SZ2L:= Za2L 40.183ft=

Right support height Za2R Z2R SZ2R := Za2R 73.29 ft=

Support height difference B2 Za2L Za2R := B2 33.106 ft=

Slope distanceC2 Aa2

2

B2

2+:=

C2 328.446 ft=

Distance to low point X02Aa2

2

H2

v2asinh

B2

2

H2

v2sinh

Aa2

2

H2

v2

+:= X02 183.251 ft=

Height at low point Z02 Za2LH2

v2cosh

X02

H2

v2

1

:= Z02 35.285ft=

Distance to wide point Xw2Aa2

2:= Xw2 163.386 ft=

Maximum conductor blowout Yw2H2

w2cosh

Xw2

H2

w2

1

:= Yw2 0.657 ft=(not including insulators)

Height at measurement point Z2 x2( ) Z02H2

v2cosh

x2 X02

H2

v2

1

+:= Z2 xc2( ) 43.418 ft=

Blowout at measurement point

Y2 x2( ) Yw2H2

w2cosh

x2 Xw2( )C2

Aa2

H2

w2

1

SY2L+ SY2L SY2R( )x2

Aa2:=

Y2 xc2( ) 0.398 ft=

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X-coordinate X2 x2( ) x2 SX2L+:= X2 xc2( ) 57.196 ft=

Rotation of coordinates:

Radius R x2( ) X2 x2( )2

Y2 x2( )2

+:= R xc2( ) 57.197 ft=

Angle x2( ) atanY2 x2( )

X2 x2( )

:= xc2( ) 0.399 deg=

Rotated angle r x2( ) x2( ) 12+:= r xc2( ) 63.834 deg=

Rotated X-coordinate Xr2 x2( ) R x2( ) cos r x2( )( ):= Xr2 xc2( ) 25.222 ft=Rotated Y-coordinate Yr2 x2( ) R x2( ) sin r x2( )( ):= Yr2 xc2( ) 51.336 ft=

Translated X-coordinate Xt2 x2( ) Xr2 x2( ) L2X+:= Xt2 xc2( ) 75.222 ft=

Translated Y-coordinate Yt2 x2( ) Yr2 x2( ) L2Y+:= Yt2 xc2( ) 1.336 ft=

Position vector cond2 x2( )

Xt2 x2( )

Yt2 x2( )

Z2 x2( )

:= cond2 xc2( )

75.222

1.336

43.418

ft=

Distance between conductors dist12 x1 x2,( ) cond2 x2( ) cond1 x1( ):=

dist12 xc1 xc2,( )

0.222

0.861

1.217

ft=

magnitude dist12 xc1 xc2,( ) 1.507 ft=

Plot the magnitude of the distance between conductors for a matrix of xc1 & xc2 distances.

N 20:= i 0 N..:= j 0 N..:= xi

.28 A1.02 A1

Ni+:= y

j.15 A2

.02 A2N

j+:=

Mi j,( )

dist12 xi

yj

,( ):= x0 70ft= xN 75ft= y0 50.312 ft= yN 57.02 ft=

M

min M( ) 1.126 ft=

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