Length of an arc between two angles: Difference between revisions

Write a method (function, procedure etc.) in your language which calculates the length of the major arc of a circle of given radius between two angles.

 

 

Illustrate the use of your method by calculating the length of the major arc of a circle of radius 10 units, between angles of 10 and 120 degrees.

 

 

=={{header|AWK}}==

# syntax: GAWK -f LENGTH_OF_AN_ARC_BETWEEN_TWO_ANGLES.AWK

# converted from PHIX

}

function abs(x) { if (x >= 0) { return x } else { return -x } }

{{out}}

<pre>

43.6332313

</pre>

 

=={{header|C}}==

{{Trans|AWK}}

#define PI 3.14159265358979323846

#define ABS(x) (x<0?-x:x)

printf("%.7f\n",arc_length(10, 10, 120));

}

{{out}}

<pre>

=={{header|C++}}==

{{trans|Kotlin}}

 

#define _USE_MATH_DEFINES

std::cout << "arc length: " << al << '\n';

return 0;

{{out}}

<pre>arc length: 43.6332</pre>

=={{header|Delphi}}==

{{Trans|AWK}}

program Length_of_an_arc;

 

Readln;

end.

{{out}}

<pre>

43.6332313

</pre>

 

=={{header|Factor}}==

 

: arc-length ( radius angle angle -- x )

- abs deg>rad 2pi swap - * ;

 

{{out}}

<pre>

43.63323129985824

</pre>

 

=={{header|Fortran}}==

The Fortran subroutine contains the MAX(DIF, 360. - DIF) operation. Other solutions presented here correspond to different interpretations of the problem. This subroutine computes the length of the major arc, which is not necessarily equal to distance traveling counter-clockwise.

* given: polar coordinates of two points on a circle of known radius

* find: length of the major arc between these points

END

 

{{out}}

<pre> first angle: 120.000000 second angle: 10.0000000 radius: 10.0000000 Length of major arc: 43.6332321

first angle: 10.0000000 second angle: 120.000000 radius: 10.0000000 Length of major arc: 43.6332321

first angle: 180.000000 second angle: 270.000000 radius: 10.0000000 Length of major arc: 47.1238899

</pre>

 

=={{header|Go}}==

{{trans|Julia}}

 

import (

func main() {

fmt.Println(arcLength(10, 10, 120))

 

{{out}}

43.63323129985823

</pre>

 

=={{header|JavaScript}}==

{{Trans|AWK}}

function arc_length(radius, angle1, angle2) {

return (360 - Math.abs(angle2 - angle1)) * Math.PI / 180 * radius;

 

console.log(arc_length(10, 10, 120).toFixed(7));

 

{{out}}

<pre>

43.6332313

</pre>

 

=={{header|Julia}}==

The task seems to be to find the distance along the circumference of the circle which is NOT swept out between the two angles.

arclength(r, angle1, angle2) = (360 - abs(angle2 - angle1)) * π/180 * r

@show arclength(10, 10, 120) # --> arclength(10, 10, 120) = 43.63323129985823

 

=={{header|Kotlin}}==

{{trans|Go}}

import kotlin.math.abs

 

val al = arcLength(10.0, 10.0, 120.0)

println("arc length: $al")

{{out}}

<pre>arc length: 43.63323129985823</pre>

 

=={{header|Perl}}==

{{trans|Raku}}

use warnings;

use utf8;

[-90, 180, 10/π],

[-90, 0, 10/π],

{{out}}

<pre>Arc length: 43.63323 Parameters: (2.0943951, 0.1745329, 10.0000000)

=={{header|Phix}}==

{{trans|Julia}}

 

=={{header|Raku}}==

degrees to radians and a postfix ᵍ to convert gradians to radians.

 

( ([-] (a2, a1).map((* + τ) % τ)) + τ ) % τ × $radius

}

\(175ᵍ, -45ᵍ, :r(10/π)) { # test gradian parameters

printf "Arc length: %8s Parameters: %s\n", arc(|$_).round(.000001), $_.raku

{{out}}

<pre>Task example: from 120° counter-clockwise to 10° with 10 unit radius

 

This REXX version handles angles (in degrees) that may be &nbsp; <big> &gt; </big> &nbsp; 360º.

parse arg radius angle1 angle2 . /*obtain optional arguments from the CL*/

if radius=='' | radius=="," then radius= 10 /*Not specified? Then use the default.*/

arcLength: procedure; parse arg r,a1,a2; #=360; return (#-abs(a1//#-a2//#)) * pi()/180 * r

/*──────────────────────────────────────────────────────────────────────────────────────*/

{{out|output|text=&nbsp; when using the default inputs:}}

<pre>

arc length = 43.6332313

</pre>

 

=={{header|Wren}}==

{{trans|Julia}}

 

 

{{out}}

<pre>

43.633231299858

</pre>

 

=={{header|zkl}}==

{{trans|Julia}}

(360.0 - (angle2 - angle1).abs()).toRad()*radius

}

{{out}}

<pre>