CODEWITHSUNDEEP
javascriptastronomy
jonsayer
jonsayer

Reputation: 93

How to calculate the latitude of the subsolar point (ie. solar declination) using javascript?

I am trying to calculate the point on Earth where the sun is directly overhead using javascript. My function for longitude works great, but I am having trouble with the latitude.

I understand that what I am looking for is called the Solar Declination, and an approximate formula for that is (source):

δ=23.45 * sin[(360/365)(284+N)]

where N is the day in the year, where January 1 is 1, Feb 1 is 32, etc.

This is the function I am using now:

    function getSolarDeclination(){
        return -23.44 * Math.sin( (360 / 365.25) * getDayOfYear() )
    }

    function getDayOfYear(){
        var now = new Date();
        var start = new Date(now.getFullYear(), 0, 0);
        var diff = now - start;
        var oneDay = 1000 * 60 * 60 * 24;
        var dayOfYear = Math.floor(diff / oneDay);
        return dayOfYear + 284
    }

However, this doesn't seem to be giving me the right number. For today, June 3, 2020, I am getting 17.607657911890687, whereas at least according to this site it should be ~22.465. This is several hundred miles off!

What am I doing wrong? Is this approximate formula just too "approximate"?

Upvotes: 2

Views: 1224

Answers (1)

John Montgomery
John Montgomery

Reputation: 7106

Looks like the formula you're using expects values in degrees, but Math.sin uses radians. If you convert it, it gives roughly the expected result:

function getSolarDeclination(){
    return -23.44 * Math.sin( (360 / 365.25) * getDayOfYear() * Math.PI/180 )
}

function getDayOfYear(){
    var now = new Date();
    var start = new Date(now.getFullYear(), 0, 0);
    var diff = now - start;
    var oneDay = 1000 * 60 * 60 * 24;
    var dayOfYear = Math.floor(diff / oneDay);
    return dayOfYear + 284
}
    
console.log(getSolarDeclination())

Upvotes: 3

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