How to calculate the distance of tow points precisely ?

Question

1.00/5 (1 vote)

See more:

C#

(float X1,float Y1,float X2,float Y2)
    {
        float dx = X1 - X2;
        float dy = Y1 - Y2;
        float sx = (float)cos( Y1 * 0.01745329252f);
        return ((float)sqrt(dx * dx * sx * sx + dy * dy) * 111195.0f);
    }

I found a implementation like this. What is its meaning ?

Posted 26-Aug-14 22:33pm

cxxxxf

Add a Solution

3 solutions

Solution 2

I can't recognize the mentioned math. The simplest way to calculate two points distance is

C++

float GetDistance(float X1,float Y1,float X2,float Y2)
    {
        float dx = X1 - X2;
        float dy = Y1 - Y2;
        return (float)sqrt(dx * dx + dy * dy);
    }

Check this Distance between two points (given their coordinates)[^] for detail

Posted 26-Aug-14 23:19pm

_Asif_

Solution 1

It's just implementing Pythagoras theorem[^]

Treat the two points as the two ends of the hypotenuse of a right angle triangle.

Posted 26-Aug-14 22:41pm

OriginalGriff

Comments

_Asif_ 27-Aug-14 5:13am

This is not a Pythagoras implementation. Pythagoras theorem is √(x^2+y^2 ). But the question is
-√(x^2 cos^2⁡ϑ+y^2) ̅

Add a Solution

Add your solution here

Treat my content as plain text, not as HTML

Preview 0

…

Existing Members

Sign in to your account

...or Join us

Download, Vote, Comment, Publish.

Your Email
Password
Forgot your password?

Your Email
This email is in use. Do you need your password?
Optional Password

I have read and agree to the Terms of Service and Privacy Policy
Please subscribe me to the CodeProject newsletters

When answering a question please:

Read the question carefully.
Understand that English isn't everyone's first language so be lenient of bad spelling and grammar.
If a question is poorly phrased then either ask for clarification, ignore it, or edit the question and fix the problem. Insults are not welcome.
Don't tell someone to read the manual. Chances are they have and don't get it. Provide an answer or move on to the next question.

Let's work to help developers, not make them feel stupid.

This content, along with any associated source code and files, is licensed under The Code Project Open License (CPOL)

nv3 · Accepted Answer · 2014-08-27T00:03:00

This is an approximation formula for the distance of two points on a sphere. The factor 0.0174... is PI / 180. That tells us that the Y values are provided in degrees. The multiplication with PI/180 converts the value to radiant.

The multiplication of dx*dx with sx*sx is the compensation for the fact that meridians run closer together in higher latitudes.

The factor 111195 corresponds to distance of one degree on the surface of the earth.

So in summary: The X und Y values are latitude and longitude in degrees and the formula delivers the distance of the two points in meters. It works only if the two points are relatively close together.