THE SHORTEST DISTANCE COVERED ALONG THE CIRCLE

MWENDA`S FORMULA.

INTRODUCTION. 

It is a formula that mwenda has come up with, that one can use to calculate the shortest distance covered along a great circle. This is where two places are located on the same latitude.it is a specific formula that one can use while calculating this distance as there is no specific formula for finding this distance.

It mainly deals with a topic in form four, longitudes and latitudes .mwenda has carefully studied this topic and that there is this specific formula that can easily be used. Mwenda has proved that this formula will be of useful when it comes to ship and plane travel.

STATEMENT OF THE PROBLEM

This formula addresses the following problems:

1, Lack of a specific  formula for calculating the distance.

2, Lack of clear understanding on longitudes and latitudes.

OBJECTIVES

To ensure there is specific formula which is mwenda`s formula.

ASSUMPTIONS

The two places are located on the same latitudes         

PROCEDURE

For instance you may be asked to calculate the shortest distance covered along a great circle or the shortest distance between two points i.e.  P(LN,XW) And Q(LN,YE) on the earth surface .Nb  L.. Mean degree north and xw means x degree west…apply the same to others.

NB.THE SHORTEST DISTSNCE BETWEEN P AND Q IS THE LENGTH OF THE ARC OF THE GREAT CIRCLE THROUGH POLE CENTER O...Centre of the earth (core).

Let x be the Centre of a circle latitude LN and k be midpoint of chord PQ.THUS

Px=Qx

Pk=Kq

<PxQ=(x+y)….angle between longitudes. Let  x+y  be g.

<PxK=g/2

Where<Pxk=(x+y)/2

Using triangle Pkx,by SOHCAHTOA.

SIN g/2=Pk/RCOSL

Pk=RCOSL .SIN g/2 where L represent latitude and g represent longitude.

Using triangle PkO

SIN @=Pk/R   =(RCOSL .SIN g/2)/R

SIN@= COSL.SIN g/2

@=2 sin~1(COSL.SIN g/2)

Angle required=2SIN~1(COSL .SIN g/2)

Distance =@/360 *2*pi*R

D= (2 SIN~1 (COSL .SIN g/2)*2*22*6370)/7*360

=111.222

2SIN~1(COSL .SIN g/2)*111.222

=111.222(2*SIN~1(COSL .SIN g/2))

D=222.444{SIN~1(COSL .SIN g/2)}

HENCE THIS IS MWENDA`S FORMULA.

HOW TO PROVE MWENDA`S FORMULA

Example one.

Advancing in mathematics form four….example is given which has similar question.

…Find the shortest distance covered between point P (65N, 25W) and Q (65N, 31E) ON THE EARTH SURFACE.

EXAMPLE TWO.

K.C.S.E 1994

A and B are two points on the latitude 40N.the two points lies on latitude 20W and 100E RESPECTIVELY.

CALCULATE…

The shortest distance from A and B along a great circle.(5mks) take R=6370km…pi=22/7.

USE MWENDA`S FORMULA TO GET THE ANSWER WITHIN ONE MINUTE. ….OR FOLLOW THE PROCEDURE.