Compass rule The compass compass or bow ditch ditch rule which which has named named after the the distinguished distinguished American navigator Nathaniel bow ditch (1773-1838) The compass compass rule is based based on the assumption assumption that that all lengths lengths wee measured measured with eual care and all angles ta!en with appro"imatel# the same precision$ %t is also assumed that the errors in measurement are accidental and that the total error in an# side of the traverse is directl# proportional to the total length of the traverse$ The compass compass rule ma# be stated as follows& follows& the correction correction to be applied applied to the latitude or departure of an# course is eual to the total closure of latitude or departure ' multiplied b# the ratio of the length of the course to the total length or perimeter of the traverse$ These correction are given b# the following euations$
1 *(d+,)
d,(d+,)
here& 1 correction to be applied to the latitude of an# course d correction to be applied to the departure of an# course * total closure in latitude or the algebraic sum of the north and south latitudes (.N*/.0,) , total closure in departure or the algebraic sum of the east and west departures (.0,/.,) d length of an# closure , total length or perimeter of the traverse
All computed corrections corrections should be a added dded to chec! whether their respective sums eual the closures in latitude and departure$ To determine the adusted latitude of an# course the latitude correction is either added to or subtracted from the computed latitude of the course$ A simple rule to remember is& if the sum of the north latitudes e"ceeds the sum of the south latitudes ' latitudes corrections are subtracted from north latitudes and added to corresponding south latitudes$ 2owever ' if the sum of the south latitudes
e"ceeds the sum of the north latitudes' the correction are applied in the opposite manner$ A similar procedure is used when adusting the departures$ After all corrections have been applied properl# to the latitudes and departures of the traverse$ it will be e"pected that a perfect closure will be obtained$
Adusted length and directions
* √ lat + de p 2
2
'
Tan 4
De p Lat '
here& *
adusted length of a course
*at
adusted latitude of a course
,ep adusted departure of a course 4
adusted hori5ontal angle between the reference meridian and a course
6A99 :;*0 (problem)
,%9TAN0
=0A:%N<
A=
>?@$8@
N @ 3 0
6;T0, *AT%T;,0 /N -9 >?3$@7
6;T0, ,0A:T;:0 /0 - >7$@3
=
8@$BC
N >B C 0
@?$@3
B1C$C3
,
8@@$>@
9 B7 38 0
3C@$@3
7?1$?
,0
1C$87
9 1C C@ 0
??B$??
C1?$@1
0D
1117$CB
9 83 >>
1C1$?B
DA
BB$8
N @@ ?
9;9
@$13 (m)
377$1? /1>B1$ C?
1>>>$> 8
a) omputing *atitudes& *at length " cos 4 b) omputing ,epartures ,ep length " sin 4
c)
,etermining Total losure in *atitudes and ,epartures .N* />?3$@7 / @?$@3 / 377$1? /1>B1$C?m (summation of north latitudes) .9* -3C@$@3 E ??B$?? E 1C1$?B - 1>>>$>8m (summation of south latitudes) .0, />7$@3 / B1C$C3 / 7?1$? / C1?$@1 /1B7$3Bm (summation of east departures) ., -111$@8 E @>1$7 - 1B@C$C8 (summation of west departure) * .N* / .9* /1>B1$C? / (-1>>>$>8) /1B$81 m
, .0, / ., /1B7$3B /
(-1B@C$C8)
/18$8m
, d1 / dC /d3 /d> /d@ / dB
111$@ 8 @>1$7 /1B7$ 3B
1B@C$C 8
d) ,etermining corrections for latitude
CL
F 1
D
16.81
F 1
5000.13
= 0.00336
A=
=
$33B
,
$33B G
>?@$8@
1$B7
G
8@$BC
C$8B
$33B
G
8@@$>@
C$88
,0
$33B
G
1C$87 3$>3
0D
$33B
G
1117$CB 3$7@
DA
$33B
G
BB$8
C$CC
96*;T%6N 20F& 1$B7/C$8B/C$88/3$>3/3$7@/C$CC 1B$81 1B$811B$81?(20F9)
e) ,etermining corrections for departure CL
F C
D
18.08
F C
5000.13
= 0.00362
A=
$3BC
G
>?@$8@
1$7?
=
$3BC
G
8@$BC
3$8
,
$3BC
G
8@@$>@
3$?
,0
$3BC
G
1C$87 3$B?
0D
$3BC
G
1117$CB >$>
DA
$3BC
G
BB$8
C$3?
96*;T%6N 20F& 1$7?/3$8/3$?/3$B?/>$>/C$3? 18$8 18$818$8(20F9) f) tabulated solutions
6;:90
6::0T%6N *AT
,0
A,H;9T0, *AT%T;,0 /N -9
A,H;9T0, ,0A:T;:0 /0 -
A= = , ,0 0D DA 9;9
1$B7 C$8B C$88 3$>3 3$7@ C$CC 1B$81
1$7? 3$8 3$? 3$B? >$> C$3? 18$8
>?1$? @87$B7 3C8$>1 1$>C 1C@$71 37>$?7 /1>@>$@ >
1>@>$@>
>@$7> B?$1@ 788$ C1@$8C
/1B@8$7 1
111>$BC @>>$? 1B@8$71
g) A,H;9T0, *AT%T;,09& A,H *AT 6;T0, *AT I
l
N6T0& 9%N0 T20 9; 6D N6:T2 *AT%T;,09 0G00,9 T20 9; 6D 96;T2 *AT%T;,09$ *AT%T;,09 6::0T%6N9 A:0 9;=T:AT0, D:6 T20 6::096N,%N< N6:T2 *AT%T;,09 AN, A,,0, T6 6::096N,%N< 96;T2 * AT%T;,09 T6 ,0T0:%N0 T20 A,H;9T0, *AT%T;,09$
A,H;9T0, ,0AT;:09& A,H ,0 6;T0, ,0 I
d
N6T0& 9%N0 T20 9; 6D 0A9T ,0A:T;:0 0G00,9 T20 9; 6D 09T ,0AT;:09' T20 ,0A:T;:0 6:0T%6N9 A:0 9;=T:AT0, D:6 T20 6::096N,%N< 0A9T ,0A:T;:09 AN, A,,0, T6 6::096N,%N< 09T ,0A:T;:09 T6 ,0T0:%N0 T20 A,H;9T0, ,0A:T;:09$
h) ,etermining the linear error of closure' bearing of the side of error and relative error of closure
*0 √ Cl ² + Cd ² *0 √ ( + 16.81 ) +(+18.08 ) ² 2
C>$B? (linear closure of error) −(+ 18.08) −CD Tan J −CL −(+16.81)
1$7@@@ J >7 @ (bearing of the side of error 9>> @) : *0+, C>$B?+@$13 1+CC$@C 9AK 1+1 (relative precision)
A,H;9T0NT9 6D *0N
*A= √ (+ 491.90 ) +(+ 45.74 ) ² 2
>?>$Cm
*= √ (+ 587.67 ) +(+609.15 ) ² 2
*, √ (−328.41 ) + ( +788.00 ) 2
8>B$Cm
2
8@3$7m
*,0 √ (−1000.42 ) +(215.82 ) ²
1C3$>3m
*0D √ (−125.71 ) +(−1114.62 ) ²
11C1$B?m
2
2
*DA √ (+374.97 ) +(−544.09) ² 2
BB$78m
b) ,etermining adusted bearing of each course TAN 4
TAN 4 A=
+ 45.74 + 491.90
N @L1?0
TAN 4 =
+ 609.15 + 587.67
N >BLC0
TAN 4 ,
+788.00 −328.41
9 B7LC30
DEP ' LAT '
TAN 4 ,0
+ 215.82 −1000.42
91CL1 0
TAN 4 0D
−1114.62 −125.71
9 83L3>
TAN 4 DA
−544.09 + 374.97
N@@LCB
) tabulated solution&
6;:90 A= = , ,0 0D DA
A,H;9T0, *0N?>$C 8>B$>C 8@3$7 1C3$>3 11C1$B? BB$78
A,H;9T0, =0A:%N< N@L1?0 N>BLC0 9B7LC30 91CL10 983L3> N@@LCB
TRANSIT RULE The method of adusting a traverse b# the transit rule similar to the method using the compass rule$ The main diMerence is that with the transit rule the latitude and departure corrections depend on the length of the latitude and departure of the course respectivel# instead of both depending on the length of the course$
1
d
lat ( Cl ) ∑ ND −∑ SL
Dep( CD ) ∑ ED −∑WD
here&
1 correction to be applied to the latitude of an# course
d
correction to be applied to the departure of an# course
* total closure in latitude or the algebraic sum of the north and south latitudes (.N* / .9*)
, total closure in departure or the algebraic sum of the east and west departures (.0, / .,) 9ince the north latitudes are positive uantities and south latitudes are negative uantities' the arithmetical sum of all latitudes is obtained if the summation of south latitudes is subtracted from the summation of north latitudes$ 9imilarl# ' the arithmetical sum of all departure is subtracted from the summation of east departures since east and west departures are positive and negative uantities' respectivel#$ Another wa# to determining these arithmetical sums is simpl# to add the absolute values of the reuired uantities$
Transit rule Dormula use to determine the latitude and departure&
A) omputing latitudes *AT *ength " os 4 *ATab >?@$8@ " os @L3 *ATbc 8@$BC " os >BLC *ATcd 8@@$>@ " os 11CLCC *AT de 1C$87 " os 1B7L3@ *AT ef 1117$CB " os CB3L>> *ATfa BB$8 " os 3>L@1
/>?3$@7 m /@?$@3 m -3C@$@3 m -??B$?? m -1C1$?B m / 377$1? m
=) omputing departures ,0 *ength " 9in 4 ,0A= >?@$8@ " ,0= 8@$BC " ,0, 8@@$>@ " ,0,0 1C$87 "
/>7$@3 /B1C$C3 /7?1$? /C1?$@1
9in @L3 9in >BLC 9in 11CLCC 9in 1B7L3@
m m m m
,00D 1117$CB ,0DA BB$8
" 9in CB3L>> " 9in 3>L@1
-111$@8 m -@>1$7 m
) Tabulated solution 6;:90
*0N
A%;T2 D:6 N6:T2
A= = , ,0 0D DA 9;9
>?@$8@ 8@$BC 8@@$>@ 1C$87 1117$CB BB$? @$13(m)
@L3 >BLC 11CLCC 1B7L3@ CB3L>> 3>L@1
6;T0, *AT%T;,0 /N -9 >?3$@7 @?$83 3C@$@3 ??B$?? 1C1$?B 377$1? /1>B1$C ? 1>>>$>8
,) ,etermining the total closure .N* />?3$@7 / @?$@3 / 377$1? / 1>B1$C? m (summation of north latitudes) .9* -3C@$@3 E ??B$?? E 1C1$?B -1>>>$>8 m (summation of south latitudes) . 0, />7$@3 / B1C$C3 / 7?1$? / C1?$@1 /1B7$3B m (summation of east departures) ., -111$@8 E @>1$7 -1B@C$C8 m (summation of west departures) * .N* / .9* / 1>B1$C? / (- 1>>>$>8)
/1B$81 m (total closure in latitude) , .0, / ., /1B7$3B / (- 1B@C$CB)
/18$8 m (total closure in departure) CL 0) ,etermining correction for latitude& F % ∑ NL −∑ SL
here& F % corrections % *AT " (F %) 9olution& 16.81
F % ( + 1461.29 ) −(−1444.48 ) $@7?
6;T0, ,0A:T;:0 /0 - >7$@3 B1C$C3 7?1$? C1B$@1 111$@8 @>1$7 1B7$3B 1B@C$C8
A= >?3$@7 = @?$@3 , 3C@$@3 ,0 ??B$?? 0D1C1$?B DA 377$1?
" $@7? C$8B m " $@7? 3$>C m " $@7? 1$88 m " $@7? @$77 m " $@7? $7 m " $@7? C$18 m
9olution chec! & sum of latitude corrections * C$8B / 3$>C / 1$88 / @$77 / $7 / C$18 1B$81 1B$81 1B$81 (chec!)
D) ,etermining corrections for departure& F C
CD ∑ ED −∑WD
here& F C corrections d *AT " (F %) 18.08
F C
+ 1670.36 −(−1652.28 ) $@>>
A= >7$@3 " $@>> />7$C7 m = B1C$C3 " $@>> /B8$? m , 7?1$? " $@>> /78B$78 m ,0 C1?$@1 " $@>> /C18$3C m 0D 111$@8 " $@>> -111B$BC m DA @>1$7 " $@>> -@>>$B@ m NOTE: since the sum of the east departures exceeds the sum of the west departures, the departures corrections are subtracted from the corresponding east departures and added to corresponding west departures to determine the adjusted departures. 9olution chec!& algebraic sum of adusted departures $ / >7$C7 / B8$? / 78B$78 / C18$3C E 111B$BC E @>>$B@ $ $$ (chec!s)
6::0T%6N
6;:90 A=
*AT C$8B
,0 $CB
A,H;9T0, *AT%T;,0 /N >?$71
-9
A,H;9T0, ,0A:T;:0 /0 - >7$C7
= , ,0 0, DA 9;9
3$>C 1$88 @$77 $7 C$18 1B$81
3$33 >$31 1$1? B$> C$?@ 18$8
@87$11 3C7$>1 1C$7B 1CC$BB 37@$1 /1>@C$ 83
1>@C$83
B8$? 78B$78 C18$3C
/1BB1$ C7
111B$BC @>>$B@ 1BB1$C7
ADJUSTMENTS OF LENGHTS AND AZIMUTHS
√ ( lat ' ) +( de p ' )² 2
A) ,etermining adusted length of each course *
√ (+ 490.71 ) + (+ 47.27 ) = 492.98 m 2
*A=
2
(adusted length of
course A=) *=
√ (+ 587.11 ) + (+ 608.90 ) =845.85 m
(adusted length of course =)
*,
√ (−327.41 ) + ( +786.78 ) =852.19 m
(adusted length of course ,)
*,0
√ (−1002.76 ) +(+ 218.32) ² =1026.25 m
(adusted length of course ,0)
*0D
√ (−122.66 ) +(−1116.62 )² =1223.34 m
(adusted length of course 0D)
2
2
2
2
2
2
√ (+ 375.01 ) +(−544.65 ) ²= 661.27 m 2
*DA
(adusted length of course DA)
Dep ' =) ,etermining adusted a5imuth of each course& TAN 4 Lat '
+ 47.27 =5 ° 30 ' TAN 4 + 490.71 (therefore a5imuth from north of course A= is @L3)
+ 608.90 TAN 4 + 587.11 >BL3 (therefore a5imuth from north of course = is >BL3)
+ 786.78 TAN 4 −327.41 B7LC> (therefore a5imuth from north of course , is B7LC>)
+ 218.32 TAN 4 −1002.76 1CL17 (therefore a5imuth from north of course ,0 is 1CL17)
−1116.62 TAN 4 −122.66 83L>> (therefore a5imuth from north of course 0D is 83L>>)
−544.65 TAN 4 + 375.01 @@LC7 (therefore a5imuth from north of course DA is @@LC7)
) TA=;*AT0, 96*;T%6N&
/>7$C7m
6;:90
A,H;9T0, *0N
A= = , ,0 0D DA
>?C$?8 m 8>@$8@ m 8@C$1? m 1CB$C@ m 11C3$3> m BB1$C7 m
A,H;9T0, A%$ D:6 N6:T2 @L3 >BL3 11CL3B 1B7L>3 C@3L>> 3>L33 C18$3C m (dep of ,0)
/B8$?( dep of =)
/78B$78 m(dep of ,)
C 8@C$1?m An(11CL3B)
-3C7$>1 m (lat of ,)
8>@$8@m
/ @87$11 m (lat of =)
D
An (>BL3)
B
1CB$C@m An(1B7L>3)
>?C$?8m An(@L3)
/ >?$71 (lat of A=)
A
-1C$7B m (lat of ,0)
BB1$C7m An(3>L33)
F
/37@$1 m (lat of
-@>>$B@ (dep of DA)
11C3$3> m'An(CB3L>>)
-111B$BC m (dep of 0D)
E
