# The Pedrom

Pedromsatsen
(The Pedrom Proposition)
Peter Mogensen
May, 2001
I. Introduction. (From a lecture to the quality class, CIEAEM conference, Set&uacute;bal, 1997.)
At the university of Karlstad, a majority of the mathematics students are future
teachers or future engineers. Generally, the students start with a single variable
calculus course. These courses are more or less identical for all students. Later,
courses may differ according to the vocation of the student.
One of the engineering programs is the Land Surveyor’s Program (Kart- och
M&auml;tningsprogrammet). Here, the students are supposed, among other subjects,
to learn about maps; how to make them and how to read them.
Until some years ago these students got a fairly traditional mathematics
education; single- and multivariable calculus and elementary linear algebra.
Nobody questioned this as few people dare to oppose a mathematician.
But one day I had a call from a teacher (non-mathematics teacher) of the Land
Surveying Program who complained arguing that their students did not get the
mathematics they needed. I was confused, of course, because the idea that
anyone should need mathematics was a new one to me. However, I asked
around among the land surveying teachers to get their view, and in the end two
major items emerged:
1&deg;. Land surveyors must be good at trigonometry.
2&deg;. Land surveyors must know that the earth is round, not flat.
With this in mind I started to revise the mathematics course on the Land
Surveying program.
The first point, trigonometry, posed few problems. Already, there are lots and
lots of excellent exercises on trigonometry in existing textbooks and from them
suitable problems could fairly easily be constructed. Here is one typical
example. The figure below shows a football goal:
C
B
D
A
P
fig 1. The penalty kick.
Given the angles APD and APB, find the angle BPC.
The second question, however, was far more difficult. How do you teach that
the earth is a round planet? You may say it and repeat it, but will the student
really internalize the knowledge in all its implications?
Fortunately, my father had once given me a problem. In my family, on my
father’s line, there are many engineers and we frequently exchange
mathematical problems of more or less realistic nature.
It was on returning from a journey in South America that my father asked me:
The aircraft started in Rio de Janeiro on the 30 th of April at 11 o’clock in the evening local time. It went
straight to Amsterdam at 10 000 metres above sea level. The speed was 840 km/h.
After how many hours of flight did I see the sun rise over the horizon?
This problem had taken me more than one night to solve and had forced me
more than once to reflect upon the shape of the earth. Of course it was far too
difficult a problem for students still believing in the flatness of our planet but it
set my mind to work. I started with something like this:
Assume that the surface of the earth is a perfect sphere with circumference 40 000 km. What is the shortest
distance from, say, Karlstad to Lisboa (going on the surface).
I would say that for the students this is an example of a true problem.
Using the formulae which Cissi and Jonas presented in their non-Euclidean
lecture yesterday it may be a simple routine task, but to the students, who
know nothing of them, it is not. I took care not to mention any spherical cosine
theorem or the like. The students must base their solution upon secondary
school trigonometry and elementary concepts from the chapter on vectors
studied one week earlier.
In the spirit of Polya we can start:
(i) Have we seen a similar problem before?
(No)
(ii) Have we learnt anything that might be useful in solving the problem?
For example could we describe the positions of Karlstad and Lisboa in a
profitable way? Karlstad is this many degrees east of Lisboa and that many
degrees north of L. How many degrees separate them along the shortest way?
There are complications here. In figure 2 we see the points A and B on the
equator 10&deg; apart. (M = centre of earth.) Now consider starting in G (fig. 3),
right north of B, and moving straight west to H, right north of A. If we know
the latitude of G and H, what is the angle GMH?
H
G
M
M
10&deg;
A
10&deg;
B
A
fig. 2
B
fig. 3
This recalls concepts from the football ”penalty kick” problem above.
Another interesting question arises as one meditates over this:
You start on the 45&deg; latitude and walk 1 km straight west. How far is it back?
You might thoughtlessly say that it is 1 km back, forgetting that the shortest
way (on the surface) is along a great circle (a circle where the centre coincides
with the centre of the earth) but the 45th latitude clearly is no great circle, so the
shortest way back is a tiny fraction of a mm shorter than 1 km. When studying
maps over small areas this will cause no problem but if the distance is longer
than a kilometre the discrepancy may be of great importance.
Here many students slowly but steadily become aware that geometry and
geography on the earth involve numerous peculiarities. We look at an ordinary
map of the world. There is a North and a South Pole. Why do we not have an
East and a West Pole as well?
fig. 4 Ordinary map of the world.
N
N
W
S
fig. 5
E
S
fig. 6 Why no East and West Pole?
What is desired is a transformation from coordinates in longitudes and
latitudes to Cartesian coordinates. Sometimes one or two students may recall
x  r cosv
the polar coordinates

y  r sinv
x  r cossin?? 
y   r??
?
Can we do the same thing in 3-dimensional space:
z  r??

  

These matters take some time to sort out in class but we finally agree upon
setting
( 0  u  180 )
u  ”angle from North Pole”
so Karlstad on 59&deg;12' north gives u K  30,8
and Lisboa (38&deg;40'N) gives u L  51, 3 .
Similarly
v  ”angle to the east of Greenwich”
which gives us
v K  13,3
( 180  v  180 &deg;)
v L   9, 2
Now, setting r = 1 (radius of earth) and exploring the depths of trigonometry
we find
x  sinu cosv 
x K  0, 4983
y   sinu sinv  so y K   0,1178  ;
z  cosu

z  0, 8590
  

 K  

x L  0,7704 
y L   0,1248
z  0, 6252 
 L  

These are unit vectors pointing from the centre of earth to Karlstad and Lisboa,
respectively. The problem is to find the angle, say  , between these vectors. 
is obtained from the scalar product:
 
cos      0,9062
 
   
 = arccos 0,9062 = 25,01&deg;
which gives the desired distance d =

360
 40 000  2780 km.
I think that it is essential that the problems are problems and not routine tasks.
The important thing is not that the students learn how to compute distances.
Should the necessity occur later in their career they will probably find a book or
a computer program to help them. The point of these exercises is that the
students must make a mental image of a spherical surface to solve them.
Therefore I have avoided a textbook. Such a book would certainly contain all
relevant formulas, and thus take away the intellectual development required to
get a complete mastery of this part of the curriculum. Instead of a book on
spherical geometry I distribute a few rather incoherent hand-written pages
which demand a good deal of patience to decode.
Most students, and nearly all successful ones, study earlier exams. It is
important that they do not get the solutions too early. These are my best
problems; if they work hard on them they are likely to learn something from it.
There is a surprisingly rich variety of problems in this area of mathematics.
Another good thing is that this part of the course comes shortly after the
chapter on planes and lines in many dimensions. Some students consider this
chapter to be rather abstract and of dubious relevance to their education so they
are pleased to see that it has a realistic dimension as well.
Sometimes as a mathematics teacher you must ask yourself whether the
students can be expected to know some facts from other disciplines required to
solve the problems. If they shall compute an integral to find the work done—
can you expect them to know that ”work equals force times distance”, just to
mention one among many examples. This issue was neatly illustrated in the
following example:
Many years ago my wife and I went to Ume&aring; in north Sweden by train. We arrived at 8 o’clock in the
morning. As we got out of the train we noted that the sun was just rising. This was on the first of
November.
Two days later we came back to G&ouml;teborg (Gothenburg) where we lived at the time. We again left the
train at 8 in the morning and I was surprised to see the sunrise this time too.
Knowing the longitudes and latitudes of G&ouml;teborg and Ume&aring;, give them for the point where the sun
stands in zenith at 8 a.m., Swedish time, in early November.
When we tried this exercise in class going to Cartesian coordinates was already
0,5227 


routine. The students fairly quickly got two vectors MG = 0,1111  and MU =
0,8453 


0,4141
0,1532  pointing from the centre of earth to G&ouml;teborg and to Ume&aring;
0,8973


respectively. But what then? It takes some consideration to realize that the
vector product will be useful here. That will provide a new vector at right angles


to both MG and MU , thus perpendicular to the plane containing M, G and U.
There were two snags, however. First the vector product of two unit vectors
need not at all be a unit vector itself so most of us first ended up with a point in
the interior of the earth. Secondly, the direction of the vector product depends




on whether you do MG  MU or if you do MU  MG so we finished the
computations in two antipodal camps, one 2000 km west of Christmas Island in
the Indian Ocean and the other close to the Mexican shores in the East Pacific.
To solve this dispute we had to rely on knowledge from outside the
mathematical world. Some were rather annoyed. ”Is this a course in astronomy
or what.”
The fact required to solve the problem, a fact I had omitted to mention in my
lectures and papers, was that on earth the sun generally rises in the east.
II. Cirklar p&aring; jordytan
Givet: En passare placeras i punkten M p&aring; Greenwichmeredianen. En cirkel
ritas p&aring; jordytan som antas vara sf&auml;risk. Cirkelns nordligaste punkt antas
hamna p&aring; latituden B grader fr&aring;n nordpolen och dess sydligaste punkt p&aring;
S&ouml;kt: Longitud och latitud f&ouml;r cirkelns &ouml;stligaste punkt.
Kommentar. Vid sm&aring; cirklar p&aring; en plan karta &auml;r problemet ej intressant:
norr
&ouml;stligaste punkt
&ouml;ster
Sv&aring;righeterna uppst&aring;r d&aring; man m&aring;ste ta h&auml;nsyn till dels jordytans kr&ouml;kning, dels att breddgraderna best&auml;ms av jordytans sk&auml;rning med parallella plan under det att l&auml;ngdgraderna
best&auml;ms av icke-parallella plan med jordaxeln som gemensam sk&auml;rningslinje. Longitudernas
avst&aring;nd fr&aring;n varandra varierar med avst&aring;ndet fr&aring;n polen och cirkelns &ouml;stligaste punkt kommer
i normalfallet inte befinna sig rakt &ouml;ster om cirkelns medelpunkt.
Thomas Martinsson har visat mig hur man traditionellt skulle angripa detta problem genom att
projicera cirkeln p&aring; ett plan som tangerar jordytan rakt ovanf&ouml;r cirkelns medelpunkt. H&auml;r
kommer en annan v&auml;g att v&auml;ljas som bygger p&aring; vektorr&auml;kning.
L&ouml;sning: Om nord- eller sydpolen ligger innanf&ouml;r cirkeln s&aring; &auml;r problemet
trivialt eftersom cirkelb&aring;gen i s&aring; fall passerar samtliga longituder. Vi bortser
d&auml;rf&ouml;r fr&aring;n detta fall, dvs vi antar att 0 ≤ B &lt; A ≤ 180.
Vi v&auml;ljer l&auml;ngdenheten s&aring; att jordradien blir 1 l.e. och placerar jorden i ett
ortonormerat koordinatsystem s&aring; att jordens centrum hamnar i origo,
ekvatorns sk&auml;rningspunkt med Greenwichmeredianen i punkten 1, 0,0 och
punkten 90&deg; &ouml;stlig l&auml;ngd 0&deg; nordlig bredd i punkten 0,1,0 . D&aring; hamnar
nordpolen i 0,0,1 .
Vektorn fr&aring;n origo, O, genom cirkelns medelpunkt (under jordytan) sk&auml;r

A  B 
cos 0 sin

2 

A  B 

jordytan i M: Vektorn OM  sin0 sin
.

2 
 A  B

cos


2
A B
 C
Vi s&auml;tter
(1)
2
sinC 
.
0
och f&aring;r OM  


cos C

Vi kan nu best&auml;mma ekvationen f&ouml;r det plan  1 som best&auml;ms av cirkeln:
x sinC  z cos C  k d&auml;r k &auml;r en konstant som kan best&auml;mmas genom att man
s&auml;tter in koordinaterna f&ouml;r tex cirkelns sydligaste punkt (sinA, 0, cosA) i
ekvationen. Denna punkt ligger i planet och satisfierar f&ouml;ljaktligen dess
ekvation:
x sinC  z cos C  sinAsin C  cosA cosC
H&ouml;gerledet kan skrivas cos(A–C). Eftersom C 
S&auml;tt
A B
 D
2
AB
A B
s&aring; A  C 
2
2
(2)
Cirkelplanets ekvation kan skrivas
x sinC  z cos C  cosD
(3)
Nu betraktar vi den longitud som tangerar cirkeln i den s&ouml;kta &ouml;stligaste
punkten. Denna longitud antas vara P grader &ouml;stlig l&auml;ngd (0 ≤ P ≤ 90).
Longituden best&auml;mmer ett storcirkelplan  2 med normalvektorn:
cos(90  P) sin90
sin(90  P) sin90  vilket efter f&ouml;renkling ger ekvationen f&ouml;r  :
2


0

xsinP  y cosP  0
(Att h&ouml;gerledet &auml;r noll f&ouml;ljer av att planet g&aring;r genom origo.)
Sk&auml;rningslinjen mellan planen (3) och (4) erh&aring;lls genom att man l&ouml;ser
ekvationssystemet
cosD

z
xtan C 

cos C


y
 xtan P  0



x

t


som ger linjens ekvation p&aring; vektorform:
(4)


x 
0

 t
y   0

 ttan P 
 

cosD  t tanC 
z 
cosC 
(5)
Den s&ouml;kta &ouml;stligaste punkten ligger p&aring; denna linje. Vi kan utnyttja att punktens
avst&aring;nd fr&aring;n origo &auml;r 1:
t 2  t 2 tan2 P  (
2
2
cosD
 t tanC)2  12
cosC
2
t (1  tan P  tan C)  2t
som ger
cosD tanC cos 2 D

1  0
2
cosC
cos C
(6)
Eftersom vi s&ouml;ker en tangeringspunkt b&ouml;r vi ha en dubbelrot och
diskriminanten &auml;r f&ouml;ljaktligen noll:
0 = 4
4
cos2 D tan2 C
–
2
cos C
cos 2 D  cos2 C  cos2 D tan 2 P  cos2 C tan2 P  cos 2 D tan 2 C  cos2 C tan 2 C
2
cos C
Efter f&ouml;renkling f&aring;s:
2
tan P 
sin2 D
2
2
cos D  cos C
Detta v&auml;rde s&auml;tts in i (6). Dessutom utnyttjar vi (1) och (2) och f&aring;r efter
omfattande f&ouml;renklingar:
t 
2sinA sinB
sinA  sinB
(7)
Detta v&auml;rde p&aring; t kan anv&auml;ndas f&ouml;r att best&auml;mma longituden och latituden f&ouml;r
den s&ouml;kta &ouml;stligaste punkten p&aring; cirkeln. Longituden sattes tidigare till P
(grader), vi kallar latituden 90–Q d&auml;r Q &auml;r vinkeln i grader fr&aring;n nordpolen
(negativ latitud svarar d&aring; mot en punkt p&aring; s&ouml;dra halvklotet). Punktens
cartesiska koordinater blir allts&aring;

x  cos P sinQ
y  sinP sinQ
z  cosQ

(8)
A B
A B
2
Ur (5) f&aring;r vi z 
A  B  t tan 2
cos
2
cos
A B
2
S&auml;tter vi dessa tv&aring; uttryck f&ouml;r z lika f&aring;s Q  arccos
A B
cos
2
s&aring; att latituden f&ouml;r den s&ouml;kta punkten &auml;r
cos
A B
2
90–Q = arcsin
A B
cos
2
cos
(9)
F&ouml;r att best&auml;mma punktens longitud P g&aring;r vi &aring;ter till (8)
x  cosP sinQ
samt till (5) och (7): x  t 
2sin A sinB
sinA  sinB
H&auml;r blir uttrycket mer komplicerat. Jag f&aring;r

A B
cos2
 2
2
sin B 1 
A
B

cos2
2  sinB
2
sinA  sinB

P  arccos



A B
2 1
A
B
2
cos
2
cos2


A

B

cos 2
2

1
2 A B

cos
2



(10)
Uttrycken (9) och (10) ger allts&aring; latitud respektive longitud f&ouml;r den &ouml;stligaste
punkten p&aring; en cirkel med centrum under Greenwichmeredianen d&auml;r A &auml;r
vinkeln fr&aring;n nordpolen f&ouml;r cirkelns sydligaste punkt och B &auml;r motsvarande
vinkel f&ouml;r cirkelns nordligaste punkt.
I och med detta &auml;r problemet l&ouml;st. Jag kan dock tillfoga en intressant kuriositet.
Betrakta specialfallet att A = 90 (grader), det vill s&auml;ga att vi har en cirkel som
tangerar ekvatorn i sin sydligaste punkt. Det visar sig att i detta fall har
longitud och latitud f&ouml;r den &ouml;stligaste punkten samma v&auml;rde — n&auml;rmare
A B
best&auml;mt arcsin cot
.
2
Till exempel den cirkel p&aring; jordytan som tangerar ekvatorn p&aring; nollmere-dianen
och har sin nordligaste punkt p&aring; den 60 nordliga breddgraden har sin &ouml;stligaste
punkt p&aring; ungef&auml;r 35,3&deg; nordlig bredd och 35,3&deg; &ouml;stlig l&auml;ngd.
Detta blir litet paradoxalt d&aring; man betraktar en cirkel som i sin sydligaste punkt
tangerar ekvatorn och passerar nordpolen i sin nordligaste (B = 0).
De flesta beh&ouml;ver en stunds eftertanke innan de inser att denna cirkel saknar
&ouml;stligaste punkt. Ju l&auml;ngre norrut man r&ouml;r sig l&auml;ngs cirkelns &ouml;stra gren, desto
l&auml;ngre &ouml;sterut kommer man tills man n&aring;r nordpolen. Men eftersom nordpolen
saknar longitud har denna cirkel ingen &ouml;stligaste punkt.
III. The Pedrom
A mathematical object, named the pedrom, is defined on the surface of the earth (considered as
an ideal sphere). For calculations vectors are used.
Given: A point A on western longitude 0 and northern latitude v
(0 ≤ v &lt; π/2).
Problem 1: To find the angle  (function of v) with the following property:
The great circle passing the latitude of A at the angle  , should have its
northernmost point on colatitude  .
By the term colatitude is meant the latitude measured from the North Pole. The problem could
be stated as follows; The great circle passing A should cross the longitude of A at the same angle
as the latitude of the same circle's northernmost point.
Solution: The earth is placed in an orthogonal system of coordinates, with its
centre in the origin, in such a way that the vector from the centre to A is
cosv 
.
A  0


sinv 
p 
Let the tangent vector of the desired great circle in A be h  q .
 
r 
cosv  p 
 q   0 . Hence p   r tanv .
It follows that 0

  
sinv  r 
The vector h and the longitude tangent in A form an angle of π/2 –  . Thus the
scalar product yields
sin  
r tan v  r tan v 
q
 0


 

r
 r

r
2
tan2 v  q 2  r2 r 2 tan2 v  r2 
2
Let s  1  tan v and the expression simplifies into
sin  
rs
(r s  q 2 )s
2
.
(1)
The great circle plane has a normal vector n which can be obtained from the
cross product
cosv  r tanv 
q sinv

 q
   r tanv sinv  r cos v 
A  h  0


 


sinv  r

q cos v

If n is normed to unit length from the origin, it touches the earth’s surface at
latitude  , above the equator. Thus, sin  =
=
q cosv
q sin v  q cos v  r tan v sin 2 v  r 2 cos2 v  2r 2 tanv sinv cosv
2
2
2
2
2
2
.
(2)
Identifying (1) and (2) gives rs  q or r(1  tan2 v )  q which can be further
simplified as
2
r  q cos v
(3)
In (1) this solves the problem:
  arctan(cos v ) .
cosv
1  cos2 v
or equivalently
(4)
NB. The author is aware that this result is more comfortably obtained using traditional
theorems from spherical trigonometry. Sometimes, however, the winding path may offer other
Definition 1. Above we have seen how each point A on the northern
Greenwich meridian defines an angle  . We call this angle the pedrom angle of
A. In its turn, the pedrom angle can be used to define a great circle, passing
through A and cutting its latitude under angle  . We call this circle a pedrom
(great) circle of A. Subsequently, the pedrom circle has a northernmost point.
We call this point a pedrom point of A. (Note that there are two pedrom points
of A, one to the east and one to the west.)
Definition 2. The set of pedrom points of the points on the northern
Greenwich meridian is called the pedrom.
In an obvious way the definition can be extended to points on the southern hemisphere and for
points on any meridian.
Problem 2: Given that A has longitude 0 and latitude v from the equator, find
the longitude t of the pedrom points.
Solution: From above we can compute a normal vector of the pedrom circle
sin v 
n  cos v  which gives the equation of the pedrom circle plane


 cosv 
x sinv  y cosv  z cosv  0
(5)
The pedrom point satisfies this equation. Moreover if the pedrom point has
longitude t, the pedrom circle crosses A’s latitude again at longitude 2t. Thus,
the point cos2tcosv ; sin2tcosv ; sinv  also satisfies equation (5):
2
cos2tcos v sinv  sin2tcos v  sinv cos v
cos2tsinv  sin2t cosv  sin v
Disregarding trivial solutions we have 2t  v    v .
Answer: The longitude of the pedrom point is
()t   /2  v .
(6)
Example: Let K be the point 15&deg;E, 60&deg;N (close to Karlstad). The northern
latitude v = 60&deg; gives  = arcsin 1/ 5  2634' , i.e. the pedrom points of K


are approximately on 15&deg;W, 63&deg;26'N and 45&deg;E, 63&deg;26'N
Remarks: Surprisingly, the pedrom circles of K have ”gitter points” inasmuch
as, 90&deg; east and west of K, they pass the 45&deg; latitude in complete accuracy.
Evidently different points on the same pedrom circle normally have different
pedrom points. For example Montreal, close to 75&deg;W,45&deg;N on the Karlstad
pedrom circle has different pedrom angle (  M  arcsin 1/ 3  3516' ),


another pedrom circle and pedrom point; 52&deg;30'W,54&deg;44'N.
Graphing the pedrom. Given v as a parameter, we can give longitude t and
latitude, say w, of the corresponding pedrom point
t 
 / 2 – v 
t
t


   
  
  
.
w 
 / 2 – arctancosv 
 / 2  arctan sint
 / 2   
Accordingly, the pedrom latitude as a function of its longitude takes the form
w (t) = arc cot sint ;
0 &lt; t ≤ π/2, π/4 ≤ w &lt; π/2.
(7)
If sketched the traditional way, this graph gives but a poor picture of the
pedrom, a sphere being required for the true graph.
Inverse pedrom. Consider a point P on the Greenwich meridian with latitude v
(π/4 ≤ v &lt; π/2). Extending the definition of pedrom point, we can ask for a
point I, such that P is a pedrom point of I. Let I have longitude x and latitude y.
From (6) above we get
 / 2  y  ()x
Choosing signs suitably we have
y   /2  x
0 &lt; x ≤π/2
which gives the set of points which have their pedrom points on the Greenwich
meridian. Such ”inverse points” have the same longitude as colatitude.
IV. Pedromsatsen
Definitioner: Universum &auml;r det euklidiska rummet av taltrippler
x, y,z d&auml;r
x, y och z reella.
Med S 2 avses sf&auml;ren
x, y,z;
2
2
x y z
2
 1.
En storcirkel &auml;r en cirkel p&aring; S 2 som ligger i ett plan som g&aring;r genom origo.
Den storcirkel som ligger i planet z = 0 kallas ekvatorn.
Punkterna N  0, 0,1 och S  0,0, 1 kallas Nord- respektive Sydpolen.
Den r&auml;ta linjen genom N och S kallas jordaxeln.
Ett plan som inneh&aring;ller jordaxeln sk&auml;r S 2 i en storcirkel. En s&aring;dan storcirkel
2
kallas en longitudcirkel. Ett plan z = a d&auml;r a  1 sk&auml;r S i en cirkel som kallas
latitudcirkel.
(En latitudcirkel &auml;r allts&aring; en storcirkel om och endast om a = 0.)
NB. Ofta skriver jag longitud och latitud d&aring; jag menar longitudcirkel respektive latitudcirkel.
Som vi ser nedan &auml;r longitud och latitud egentligen vinklar. F&ouml;rhoppningsvis orsakar detta inte
problem.
L&aring;t en godtycklig punkt a ,b ,c  p&aring; S vara given. Vinkeln   arccosc kallas
punktens colatitud. Punktens latitud &auml;r vinkeln   arcsinc . Vidare kan man
finna en vinkel  mellan –π och +π s&aring;dan att a  cos  sin  och
b  sin  sin  . Denna vinkel  kallas punktens longitud.
2
Anm. Longituden &auml;r allts&aring; inte entydigt best&auml;md d&aring; vinkeln &auml;r &plusmn;π. M&aring;nga f&ouml;redrar att
definiera longituden p&aring; intervallet [ 0 , 2π [.
M&auml;ngden av de punkter p&aring; den longitudcirkel som g&aring;r genom (1,0,0) och
vilkas x-koordinat &auml;r positiv kallas Greenwichmeridianen. Punkter p&aring; S 2 med
positiv y-koordinat s&auml;gs ligga &ouml;ster om Greenwich, punkter med positiv zkoordinat s&auml;gs ligga norr om ekvatorn. De v&auml;stra och s&ouml;dra hemisf&auml;rerna
definieras analogt.
Sats (Pedromsatsen):
F&ouml;ruts&auml;ttningar: En sf&auml;risk cirkelskiva C 1 p&aring; S 2 har sin nordligaste och
sydligaste punkt p&aring; Greenwichmeridianen. I den sydligaste punkten tangerar
C 1 ekvatorn. L&aring;t E vara den &ouml;stligaste punkten p&aring; C 1 . En storcirkel C P dras

  i E mellan C P och longitudcirkeln genom
2
E &auml;r lika med latituden f&ouml;r den nordligaste punkten p&aring; C P . Kalla den s&aring;lunda
best&auml;mda nordligaste punkten p&aring; C P f&ouml;r P.
(P &auml;r allts&aring; pedrompunkt f&ouml;r E.)
genom E p&aring; s&aring; s&auml;tt att vinkeln
P&aring;st&aring;ende: P ligger p&aring; 90&deg; &ouml;stlig longitud fr&aring;n Greenwichmeridianen.
Bevis: L&aring;t C 1 har medelpunkt M (p&aring; S 2 ) och radie r. Den sf&auml;riska triangeln
EMN har d&aring; en r&auml;t vinkel i E.
MN 

2
 r och ME  r . Pythagoras’ sats ger cos NE  tan r
Detta ger att E har colatitud arccostanr och f&ouml;ljaktligen latitud arcsintanr
Vinkeln ENM f&aring;s ur Napiers formler:
sinENM 
sinEM
 tanr vilket ger att
sinMN
vinkeln ENM = arcsintanr
(1)
(D&auml;rmed &auml;r det &auml;ven visat att punkten E har samma longitud som latitud.) Vi
skall nu best&auml;mma vinkeln ENP f&ouml;r att f&aring; longituden f&ouml;r P. Triangeln ENP har
en r&auml;t vinkel i P s&aring; Napiers formler ger


cosENP  tanNP cotEN  tan NP cot  ENM  eller
2
cosENP  tanNP tan ENM
samt
cosNEP  sinENP cosNP
(2)
Enligt f&ouml;ruts&auml;ttningarna kan detta skrivas
sinENP  tanNP
(3)
Nu delas (2) ledvis med (3):
cotENP  tanENM vilket genast ger att summan av vinklarna
ENP och ENM &auml;r en r&auml;t. Av detta f&ouml;ljer p&aring;st&aring;endet.
Q.E.D.
F&ouml;ljdsats: L&aring;t P vara en punkt norr om den 45 nordliga breddgraden. En
storcirkel C P dras som tangerar latituden i P. Kalla sk&auml;rningspunkten mellan
C P och ekvatorn f&ouml;r Q.
F&ouml;lj C P fr&aring;n P mot Q intill dess att vi n&aring;r en punkt som &auml;r lika m&aring;nga l&auml;ngdsom breddgrader fr&aring;n Q. Kalla den punkten f&ouml;r E. D&aring; g&auml;ller att den nordliga
latituden f&ouml;r P &auml;r samma vinkel som den vinkel C P bildar med longituden
genom E.
Bevis: P&aring;st&aring;endet f&ouml;ljer direkt ur Pedromsatsen om man noterar dels att Q av
symmetrisk&auml;l m&aring;ste ligga 90 longitudgrader fr&aring;n P, dels att det finns en punkt
som uppfyller det givna villkoret f&ouml;r E s&aring; snart P ligger norr om 45&deg; nord. Detta
inses av (7) i avsnitt III.
Q.E.D.