Forum Discussion
QuickBaseCoachD
6 years agoQrew Captain
You can download free excels dumps of every zip code
http://federalgovernmentzipcodes.us/
That will allow you to geocode your zips.
Then you can wake up sleeping Pythagoras from 2,500 year ago who will answer your question.
// In the USA, the typical distance between integer Longitude lines is about 53 miles. (these are the east / west coordinate)
// They are further apart at the equator (69 miles) and approach zero at the North Pole.
// but we have few courses at the north pole or the equator, so lets just call it 53.
// In the whole world, its 69 miles between integer Latitude lines.
// With credit to Pythagoras, we know that for a right angle triangle A^2 + B^2 = C^2. I will use (A^2 for A squared)
// We want to find the length of the C diagonal where A is the North South distance and B is the West East Distance
// Let LA1 be the LAtitude 1
// Let LO1 be the LOngitude 2
// Let LA2 be the LAtitude 1
// Let LA2 be the LOngitude 2
// So C^2= A^2 + B^2
// C = SQRT (A^2 + B^2)
// C = SQRT ((69*(LA1-LA2))^2 + (53*(LO1-LO2))^2)
// note that to take a square root you raise it to the power of 1/2 or 0.5
var number OriginLat = [Patient Zip Code - Latitude];
var number OriginLong = [Patient Zip Code - Longitude];
var number DestLat = [Consultant Zip Code - Latitude];
var number DestLong = [Consultant Zip Code - Longitude];
var number Distance =
Round(
((69*($OriginLat - $DestLat))^2 + (53*($OriginLong - $DestLong))^2)^0.5
);
If($OriginLat =0 or $DestLat=0,0,$Distance)
http://federalgovernmentzipcodes.us/
That will allow you to geocode your zips.
Then you can wake up sleeping Pythagoras from 2,500 year ago who will answer your question.
// In the USA, the typical distance between integer Longitude lines is about 53 miles. (these are the east / west coordinate)
// They are further apart at the equator (69 miles) and approach zero at the North Pole.
// but we have few courses at the north pole or the equator, so lets just call it 53.
// In the whole world, its 69 miles between integer Latitude lines.
// With credit to Pythagoras, we know that for a right angle triangle A^2 + B^2 = C^2. I will use (A^2 for A squared)
// We want to find the length of the C diagonal where A is the North South distance and B is the West East Distance
// Let LA1 be the LAtitude 1
// Let LO1 be the LOngitude 2
// Let LA2 be the LAtitude 1
// Let LA2 be the LOngitude 2
// So C^2= A^2 + B^2
// C = SQRT (A^2 + B^2)
// C = SQRT ((69*(LA1-LA2))^2 + (53*(LO1-LO2))^2)
// note that to take a square root you raise it to the power of 1/2 or 0.5
var number OriginLat = [Patient Zip Code - Latitude];
var number OriginLong = [Patient Zip Code - Longitude];
var number DestLat = [Consultant Zip Code - Latitude];
var number DestLong = [Consultant Zip Code - Longitude];
var number Distance =
Round(
((69*($OriginLat - $DestLat))^2 + (53*($OriginLong - $DestLong))^2)^0.5
);
If($OriginLat =0 or $DestLat=0,0,$Distance)
SarahBurres
6 years agoQrew Trainee
I have attempted this but I must have made a mistake. The miles should be 3 and I am getting 5000.
------------------------------
Sarah Burres
------------------------------
var number OriginLat = [LatitudeVendor];
var number OriginLong = [LongitudeVendor];
var number DestLat = [LatitudePlant];
var number DestLong = [LongitudePlant];
var number Distance = Round(((69*($OriginLat - $DestLat))^2 + (53*($OriginLong - $DestLong))^2)^0.5);
If($OriginLat =0 or $DestLat=0,0,$Distance)
var number OriginLong = [LongitudeVendor];
var number DestLat = [LatitudePlant];
var number DestLong = [LongitudePlant];
var number Distance = Round(((69*($OriginLat - $DestLat))^2 + (53*($OriginLong - $DestLong))^2)^0.5);
If($OriginLat =0 or $DestLat=0,0,$Distance)
------------------------------
Sarah Burres
------------------------------