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In this particular type of projection, the surface of the projected plane
makes up a cone that circumscribes the globe with a selected parallel
line tangential to the globe as seen in Figure A. Such a line is defined
as the standard parallel with the characteristic that it is the arc of
a circle with its centre at the vertex of the cone and with a true length.
The true length of the standard parallel can be obtained as follows:
(refering to Figure B) the cone LAT is seen to be tangential to the sphere
at point L intersecting with the latitude LT with angular distance <LOB.
Since the plane is conic in shape and is tangential along all point on
this latitude, this implies that the true length of the parallel is equal
to the circumference of a circle with radius LC, which is equal to the
radius of the circle * cos (angular distance of the latitude). This is
because the latitude LT is by definition parallel to the equator BD, therefore,
<BOL = <OLA. In addition, <LCO is a right angle because all parallels
intersect the polar axis perpendicularly. Using trignometric functions,
cos(<CLO)=LB/LO. Given that <CLO is the angular distance and LO
is the radius of the circle, LB is equal to the circle's radius * cos
(angular distance of the latitude). Therefore, the true length of the
parallel is equal to 2 * 3.141592654...* circle's radius * cos (angular
distance of latitude).
Other parallels will appear as concentric arcs with the standard parallels.
In fact, there are three different situations that can be used to project
the other parallels. The first situation is depicted in Figure D where
the projection of the latitude is rotating a line of length KA at point
A (vertex of the cone) such that LM =KL. The length KA is equal to the
r * cot (angular distance of standard parallel) - square root {[r * cos
(standard parallel) - r * cos (other parallel)] to the power of 2 + [r
* sin (other parallel) - r * sin (standard parallel)] to the power of
2} where r is the circle's radius. This conclusion is reached by using
the right angle triangle LMP to calculate a length for LM. By drawing
a perpendicular line MP, its distance is equivalent to the difference
between the vertical distance of the two latitudes. As such MP = r*sin(latitude)-r*sin(standard
parallel). The distance LP is the difference between the horizontal distance
of the two latitudes. As such LP = r*cos(standard parallel) - r*cos(latitude).
Therefore by pythagorus, LM = square root {[r * cos (standard parallel)
- r * cos (other parallel)] to the power of 2 + [r * sin (other parallel)
- r * sin (standard parallel)] to the power of 2}. With LA = r*cot(standard
parallel), KA becomes r * cot (angular distance of standard parallel)
- square root {[r * cos (standard parallel) - r * cos (other parallel)]
to the power of 2 + [r * sin (other parallel) - r * sin (standard parallel)]
to the power of 2}. With the projected meridians, it produces the map
below it in Figure E.
The second situation that occurs is depicted in Figure F, in that the
line rotated around A is obtained by extending the line OM to LA. As such
the distance KA is equal to r* cot (standard parallel) - r*tan (latitude-standard
parallel). This is because LO is perpendicular to LA (based on the satisfaction
of two of the basic requirements - touching at a tangent point and going
through the centre of the circle), therefore triangle KLO is a right angle
triangle with <OLK = 90°. <KOL = <MOB - <LOB or <KOL
is equal to the difference between the angular distance of the latitude
with the standard parallel. Therefore using trignometric relations and
the fact that LO is equal to the radius of the circle, LK is equal to
r*tan (latitude-standard parallel). Therefore, AK is equal to r* cot (standard
parallel) - r*tan (latitude-standard parallel). The map that this projection
produces is illustrated in Figure G.
The last possible projection is that seen in Figure H in that the distance
AK is created by drawing a line segment from KM to LA such that they are
perpendicular to each other. Clearly LO = MO since they are both the radius
of the circle. As a result the angles <OLM and <LMO, which are opposite
to equal sides, are also equal. Since all the angles in a triangle add
up to 180° and <LMO= latitude - standard parallel (as shown in
the situation of Figure F). <OLM = <LMO = (180° - latitude +
standard parallel)/2. Since <OLA is 90°, this implies that <KLM
= 90 - <MLO therefore <KLM = (latitude - standard parallel)/2. Given
that <MKL is 90° and LM is equal to square root {[r * cos (standard
parallel) - r * cos (other parallel)] to the power of 2 + [r * sin (other
parallel) - r * sin (standard parallel)] to the power of 2}(obtained from
the first situation in Figure D), this implies that LK is square root
{[r * cos (standard parallel) - r * cos (other parallel)] to the power
of 2 + [r * sin (other parallel) - r * sin (standard parallel)] * cos
[(latitude - standard parallel)/2] and KA is r*cot (standard parallel)
- LK. Its map is seen below in Figure I.
The fact that LA is equal to the radius of the circle * the cot ( the
standard parallel) can be derived at using Figure B.
- Since LT and BD are parallel lines, this implies that <TLO = <LOB
= the angular distance of the latitude.
- <ALO = 90° because OL goes through the centre and intersects
a tangent line. Therefore <ALT + <OLT = 90°.
- <ACL = 90° because all parallels intersect polar axis perpendicularly.
- By the interior angle sum theroem, <LAC + <ACL + <CLA = 180°
and <CLO + <LOC + <OCL = 180°
- Since <LCO = <ACL = 90°, this implies that <CLO + <LOC
= <LAC + <CLA = 90°. From before, one already knows that <CLO
+ <ALC = 90° and <LAC + < CLA= 90° therefore, <LAC
= <OLC = angular distance of latitude.
- tan (<LAC) = OL/LA and since OL is the radius of the circle therfore
LA = circle's radius * cot (angular distance of latitude).
On the other hand, the projected meridians at 1° intervals will divide
this arc into 360 equal arcs by appearing as straight lines radiating
from the vertex at equal angles (similar to Figure C). The angle at the
vertex (A in Figure C) from which the meridians will divide equally is
obtained using the following ratio:
- <LAT/ 360 = arcLT/circle with radius LA
- <LAT/360 = 2 * 3.141592654 * r * cos (standard parallel)/2*3.141592654
* radius * cot (standard parallel)
- <LAT = 360 sin (standard parallel)
Therefore, with the following information provided conical map projection
such as those seen in Figure E, G, and I are produced.
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