egk%mica.Berkeley.EDU@ucbvax.berkeley.edu (10/10/89)
I just loaded my Bellcore V&H Coordinate tape onto my VAX to help with my project I mentioned in an earlier post, my new question is: What are the units of those coordinates? and What document am I missing that would tell me this? Thanks! E+ 0 + @ + @ + \ * | * / % + -- EGK -- + % / * | * \ + @ + @ + 0
eric@cirr.com (Eric Schnoebelen) (10/13/89)
[ I started to mail this, then thought it might be of general interest.
Enjoy, Eric ]
In article <telecom-v09i0440m07@vector.dallas.tx.us> you write:
-I just loaded my Bellcore V&H Coordinate tape onto my VAX to help with
-my project I mentioned in an earlier post, my new question is:
-What are the units of those coordinates? and What document am I
-missing that would tell me this?
Well, it is my understanding that the units in the V&H master database
are miles. This allows the milage based costing of the phone company
services to be calculated using the simple distance formula:
sqrt( (x1- x2)^2 + (y1 - y2)^2 )
although the phone companies really use the following formula:
sqrt( ( (v1 - v2)^2 + (h1 -h2)^2 )/10 ).
Don't ask me why.... I once asked the resident network design
engineers, and they gave me an answer that I don't really remember,
but could be summed up as they really didn't know either....
Hope this helps,
Eric Schnoebelen eric@egsner.cirr.com
"My other computer is a Convex" schnoebe@convex.comde@cs.rochester.edu (Dave Esan) (11/15/89)
In article <telecom-v09i0445m03@vector.dallas.tx.us> you write: >Well, it is my understanding that the units in the V&H master database >are miles. This allows the milage based costing of the phone company >services to be calculated using the simple distance formula: >sqrt( (x1- x2)^2 + (y1 - y2)^2 ) >although the phone companies really use the following formula: >sqrt( ( (v1 - v2)^2 + (h1 -h2)^2 )/10 ). V and H coordinates are points on a grid spread over North America. Being points they will be dimensionless. One can calculate the distance between any two points using the V and the H coordinates and simple geometry - eg the distance is sqrt ( (x1- x2)^2 + (y1 - y2)^2 ), as noted above. But this will be wrong. The Earth is round, and this distance will not be correct. There is a distance method given in FCC #10, page 13. Basically it is as follows: 1. Calculate the difference in V coordinates, and H coordinates. 2. Divide each by three. 3. Square the numbers and add them. 4. If the sum of the square is > 1777 go to step #2. (Forgive me for using a goto statement.) 5. If the sum of the square is < 1777 multiply it by a fudge factor based on the number of divisions done. 6. Take the square root of the product, and round up. Of course, if it is zoned city you have to worry if the distance is < 40 miles, in which case you have to recalculate using the zones coordinates rather than the master coordinates. Unless of course, this distance is greater than 40 miles, in which case you use the regular calculation. The distance between coordinates is supposed to be about 180 yards. (Maybe 173 yards = 1/10 mile?). The point (0,0) is someplace in the Atlantic. Unlike ATT which calculates that cost of a call to Puerto Rico based on 3 costing bands, Sprint calculates the mileage using a fake coordinate that has a negative H component. --> David Esan rochester!moscom!de
de@cs.rochester.edu (Dave Esan) (11/28/89)
This followup to an article bounced back here. I'll try again. In article <telecom-v09i0445m03@vector.dallas.tx.us> you write: >Well, it is my understanding that the units in the V&H master database >are miles. This allows the milage based costing of the phone company >services to be calculated using the simple distance formula: >sqrt( (x1- x2)^2 + (y1 - y2)^2 ) >although the phone companies really use the following formula: >sqrt( ( (v1 - v2)^2 + (h1 -h2)^2 )/10 ). V and H coordinates are points on a grid spread over North America. Being points they will be dimensionless. One can calculate the distance between any two points using the V and the H coordinates and simple geometry - eg the distance is sqrt( (x1- x2)^2 + (y1 - y2)^2 ), as noted above. But this will be wrong. The Earth is round, and this distance will not be correct. There is a distance method given in FCC #10, page 13. Basically it is as follows: 1. Calculate the difference in V coordinates, and H coordinates. 2. Divide each by three. 3. Square the numbers and add them. 4. If the sum of the square is > 1777 go to step #2. (Forgive me for using a goto statement.) 5. If the sum of the square is < 1777 multiply it by a fudge factor based on the number of divisions done. 6. Take the square root of the product, and round up. Of course, if it is zoned city you have to worry if the distance is < 40 miles, in which case you have to recalculate using the zones coordinates rather than the master coordinates. Unless of course, this distance is greater than 40 miles, in which case you use the regular calculation. The distance between coordinates is supposed to be about 180 yards. (Maybe 173 yards = 1/10 mile?). The point (0,0) is someplace in the Atlantic. Unlike ATT which calculates that cost of a call to Puerto Rico based on 3 costing bands, Sprint calculates the mileage using a fake coordinate that has a negative H component. --> David Esan rochester!moscom!de