The answer is a formula that converts the source number into a decimal that Excel can evaluate as a time value. In the example sheet, I've entered =A2/24 into C2 and copied the formula to cells C3:C5. In my engineering work, I need to display the distance along a roadway curve as follows: 2537.43 feet along the arc from the beginning will be displayed as 23+37.43. Is there a way to format a cell so that if I type in, say, 913.28 and hit enter, the cell displays '9+13.28.
This site is owned and maintained by Wight Hat Ltd. ©2003-2018.Our full terms & conditions can be found by.Whilst every effort has been made to ensure the accuracy of the metric calculators and charts given on this site, we cannot make a guarantee or be held responsible for any errors that have been made. If you spot an error on this site, we would be grateful if you could report it to us by using the contact link at the top of this page and we will endeavour to correct it as soon as possible.this page last updated:: Sun 22 Jul 2018.
Getting StartedFor those who are in a rush for the solution and don't need all thebackground information, jump to. Or download the calculation workbook, and enter your longitude and latitude.The search for an accurate solution to this problem has led me tonumerous sites and attempted solutions.
A long list of related sitesis included at the end of all of this, but the most crucial to whatI've found to be the current end of the road are these:along withAccuracy is ImportantA formula that is accepted to provide results that are accurate towithin millimeters is known as Vincenty's formula. Naturally the accuracyof the results depends in large part on the accuracy of the latitude/longitudepairs describing the two points.Why all the fuss over accuracy? Well, from what I've seen of otherformulas, especially those written as a single worksheet function,their values differ quite a bit for what might be considered 'lifecritical' situations. Typically they are short by some number of meters,typically about 20 to 30 feet per statute mile, and after flying just30 or 40 miles, I wouldn't care to land several hundred feet shortof the approach end of a runway, much less be off by over 7 mileson a trip between Los Angeles and Honolulu.Since the general tendency in dealing with these types of calculationsis to 'measure it with a micrometer, mark it with chalk and cut itwith an axe', what we have here is a very precise micrometer to beginthe measuring process with.