Creating degree values from numbers

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I am doing some work with the local fire brigade, instructing the members on the use of the GPS re Bearings & Headings.
I know my sine, cos and tan functions, using standard tables.
My questions are:
How do I use the Scientific calculator in Windows to convert a number to a degree value? I have done it previously but I am in a muddle.
How do I use the Function attribute in Excel to convert a number to a degree value? The sine function also made no sense to me. I don’t want radians.
In fact I would like to create my own tables in Excel to convert calculated numbers (be they derived from sine, cos, or tan) to a degree value. Perhaps someone would like to post me a worksheet.
Bruno.

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There are 360 degrees in one circle. There are 2*pi radians in circle. Thus
Degrees = Radians / (2*PI()) *360 or

Degrees = 180 * Radians / PI()

Steve

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There are 360 degrees in one circle. There are 2*pi radians in circle. Thus
Degrees = Radians / (2*PI()) *360 or

Degrees = 180 * Radians / PI()

Steve

Within Excel of course, you can use the formula =DEGREES(A1) to convert A1 in radians into Degrees
=RADIANS(B1) does the opposite conversion.

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Dear Steve, I am aware of the radians and the formula posted above – that is not what I am after.
Please see attachment.
Say I know two co-ordinates (A & B), that gives me the a distance between the two co-ordinates (AB) and my sine value is 0.853764855 this will equate to 58.6 degrees.
I can therefore inform the fire fighters to proceed in a Heading of (in this case) 58.6 + 270 degrees = 329 degrees to go from co-ordinate A to co-ordinate B.
What I would like to know is how to use Windows Scientific calculator to get 58.6 degrees from a calculated value of 0.853764855
What I also would like to know is how to enter that calculated number (0.853764855) into an Excel worksheet and derive at 58.6 degrees. Like I stated: I have tried the Sine function facility but it made no sense to me.
Bruno

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Assume your sine value (0.853764855) is a calculation in cell E5.
Try this formula in some other cell ….
=DEGREES(ASIN(E5))

Tim

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Ripper, thank you very much for that. I hope that I will be on my way using cos and tan values also.
Bruno.

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If we can help on cos and tan values, please post some examples as you did with sine.

Tim

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Thank you Tim. I used your example to create cos & tan cells. Working with Eastings & Northings, one often obtains negative values. So in effect I still have to sketch a small drawing like the one above to determine the Bearing that the fire fighters have to take, because sometimes the angles are added to the cardinal points and sometimes subtracted from.

The only other thing I would like to know is how to use Windows Scientific calculator to get 58.6 degrees from a calculated value of 0.853764855
It annoys me greatly that I have used this program previously but am now not able to make it work for me – I should have kept my slide rule (do they still make them?).
Bruno with thanks to all who responded.

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Working with Eastings & Northings, one often obtains negative values. So in effect I still have to sketch a small drawing like the one above to determine the Bearing that the fire fighters have to take, because sometimes the angles are added to the cardinal points and sometimes subtracted from.

Have you tried a conditional formula such as (=IF) to add/subtract to/from cardinal point? If not, can you provide a few examples of what you need?

Tim

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The conversion is not a radian to degree issue. It is just the arcsine issue (or “inverse sine”) [What is the angle whose sine is…]

Make sure Decimal and Degrees are selected (they seem to be the default)

Enter the value 0.8537648555 [or calculate 3842/4500.1]

“Check” the Inv (for Inverse) box

Press [sin] button

It will display 58.623….

Steve

PS Tim already supplied the Excel answer:
=DEGREES(ASIN(3842/4500.1))

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Thank you Steve, this works. I must have forgotten about the inverse box. Please tell me what the box next to it is, the one that states “hyp”, as I seem to have forgotten that function also.
Bruno

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these sites may be useful ….

http://en.wikipedia.org/wiki/Hyperbolic_function

http://tipsforspreadsheets.com/microsoft_excel_2003_function_0024.html

http://office.microsoft.com/en-us/excel/HP052012531033.aspx

Tim

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Dear Tim & others.

Please see attached file with included commends. Have a look at the muddle of formulae in range I3:K6 if this can be tidyied up a bit to give automatic (correct) answers I would appreciate it.

Bruno

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If I understand correctly, it seems to me that the bearing (in degrees) is:
=180-G5+SIGN(B5)*90

Steve

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Nice try Steve. Yes in this case you obtained the correct Bearing.
However (using the same given worksheet) transpose the Eastings so that cell B2 has a value of 398634 and cell B3 has a value of 400977 and see what your answer is. The answer should in this case be 58.62 degrees.
You can of course play with transposing the Northings and this should give degree readings of 148.62 or 211.38 depending on your Easings (transposed or otherwise).
I noticed that you had “sign” (B5) – why not also (or somewhere) “sign” C5?
Back to the drawing board.
Bruno

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I have attached a file with 4 examples of the data. I am sure we can figure this out.
What are the correct answers for tab named Example 3 and tab named Example 4?

Briefly, what determines correct answer for each example?

Tim

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Dear Tim and Steve
Attached please find the response to Tim’s quesion.
I have written a small note in A20 (sheet 2, 3, and 4) to explain the basics of UTM co-ordinates – nothing fancy and easy enough for you guys to work out.
Let’s have a look what you can come up with.

Bruno.

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If you transpose the Eastings, shouldn’t the bearing be <45°? It seems to me that the bearing is North of NE line of 45, not East of NE as your 58.6 indicates….

The 58.6 angle is the angle from East direction up to the Line. But you want the bearing down from the North (0) to the East which is 90-58.6 = 31.4 which is what my calculation should give.

Steve

PS Another way to look at it. If you transpose the Eastings you just change the E/W direction. The original example's bearing is 328, which is 32° South of North (360, or 58°North of West angle of 270°). By transposing the eastings you still remain 32° South of North and you are now on the East half of the compass and thus 58°North of East angle of 90°).

PPS. The tan will give a Div/0 error when the eastings are the same, which means your bearing is either 0 (due N) or 180 (due S). To avoid this error, in the calc use:
=IF(B5=0,90+SIGN(C5)*90,180-G5+SIGN(B5)*90)

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Sorry Steve/Tim. Steve you are absolutely correct re worksheet 2, in that by transposing the Easings value, you indeed have 31.38 degrees for the actual Bearing line. My humble apologies – I was too fixed on the actual angle within the ABC triangle without looking at the Bigger Picture by taking that away from 90 degrees as I have done in cell J4.

In the attached spreadsheet (worksheet 2), I have placed a .jpg sketch of the whole situation to give everyone a clear idea of UTM co-ordinates and the various angles involved.

I have also attached the same sketch (not to scale) as a separate file.
Bruno.

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Hello Steve/Bruno – This has been a most interesting and informative project. Bruno’s knowledge and illustrations are terrific. Steve’s logic and explanations are on point. I enjoy this kind of challenge. You both know the subject very well.

I have another question ….
Steve’s formula gets the same answer as Bruno for examples 1,2, and 4. What causes the differences in conclusions for example 3 (211.38 vs. 238.62)?

Tim

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Dear Tim/Steve
I have tried Steve’s formula =IF(B5=0,90+SIGN(C5)*90,180-G5+SIGN(B5)*90)
and this works except in example three where it shows 238 degrees rather than 211 degrees. Why this is I have no idea and that is something that Steve may like to address with a modified formula.
As you can see, these formulae will need to be “bullet proof” as lives may depend on the interpretation by the fire fighters to head onto a correct Bearing line to get to the seat of the fire/people.
I know that I am out of my depth here, so that is why I posted this thread to see if an answer can be generated.
Bruno.

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It seems to work for me. The active part of the formula
=IF(B5=0,90+SIGN(C5)*90,180-G5+SIGN(B5)*90)

That is not zero is:
=180-G5+SIGN(B5)*90
= 180-(58.62) + (+1)(90)
= 270 – 58.62
= 211.38

The logic of the formula is:
1) If B5 = 0 then the there is no difference in the eastings adn the line is vertical (and the tan will be undefined, so can not be used to calculate.
The portion: 90+SIGN(C5)*90
has 2 values 90-90 = 0 (bearing North) or 90+90 = 180 (South) which is determined by the sign of C5. C5 is negative when B is North of A and positive when B is South of A.

2) when the line is NOT vertical (B50) then the bearing can be determined by the angle derived from tangent (G5) and the sign of the Eastings (B5).

The Easting value defines which have of the compass we are on: East Half or West half. If B5 is negative, the angle points Easterly and 90 (180-90, the East direction) will be used. If B5 is positive, the angle points Westerly and 270 ((180+90, the West direction) will be used.

The arctangent angle is positive when the slope is positive which occurs when angle is in the NE quadrant and the SW Quadrant. So in those cases you will get (NE) 90-angle or (SW) 270-angle. The arctangent is negative when you are in the SE quadrant and the NW quadrant. In these cases you get 90-(“-angle”) = 90+angle or 270+angle.

The acrtan angle is exactly 0 when you are in the East or West direactions which will give either 90 (E) or 270 (W) based on whether B5 is pos or neg

Does this clear it up any?

Steve

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It seems to work for me. The active part of the formula
=IF(B5=0,90+SIGN(C5)*90,180-G5+SIGN(B5)*90)

That is not zero is:
=180-G5+SIGN(B5)*90
= 180-(58.62) + (+1)(90)
= 270 – 58.62
= 211.38

Hi Steve/Bruno – I think the problem came about when file named “UTM co-ordinates rev 3.xls” was created. On the tab named “Example 3” cell G4 got changed to =B5/C5. I think it should be =C5/B5. If that is correct, all should be OK now.

Tim

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I suggest you look into the ATAN2 function. See http://en.wikipedia.org/wiki/Atan2

What you’re trying to do is convert from rectangular or Cartesian coordinates to polar coordinates ( http://en.wikipedia.org/wiki/Polar_coordinates ). As you’ve already figured out, the distance or radius is calculated using the Pythagorean theorem to calculate the hypotenuse of a right-angle triangle. For the theta value (degrees of rotation from the origin), the best way to calculate it is using ATAN2, which calculates the arc tangent and figures out all the sign issues that you’re trying to work out using conditionals. In Excel, ATAN2(X; Y) gives the theta value in radians for a given displacement in the X and Y directions, for the standard Cartesian coordinate system. However, standard polar coordinates rotate counter-clockwise, with 0 degrees corresponding to (1, 0) – i.e. an eastern heading. Compass bearings (or azimuth) transpose this, using a clockwise increase in degrees with 0 being north. So, when using ATAN2, we transpose our X and Y to get the transposed polar coordinate system. So the difference in Northings becomes our X, and the difference in Eastings becomes our Y. It’s also very important to get the right sign in our differences, so that ATAN2 can calculate the correct quadrant: we’re going from point A to point B, so we have to calculate location B’s Northings minus location A’s, and likewise for the Eastings. That gives us this formula, using the cells you defined in the spreadsheet with location A in (B2, C2) and location B in (B3, C3):

=DEGREES(ATAN2(C3-C2; B3-B2))

There’s one missing piece of the puzzle, though, as this gives a result of -180 to +180 degrees, and we really want 0 to 360. So, we need to add 360 to the result when the difference in Eastings is negative. One slick way of doing this is to add 180 – 180*SIGN(B3-B2) to the result. If the sign is positive, the two numbers cancel out, but if it’s negative they add up to 360. The full formula would then be:

=DEGREES(ATAN2(C3-C2; B3-B2))+180-180*SIGN(B3-B2)

Give this a try with several sets of locations to confirm that it’s consistently giving you the right bearing, but I believe this is the formula you’re looking for.

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I suggest you look into the ATAN2 function. See http://en.wikipedia.org/wiki/Atan2

=DEGREES(ATAN2(C3-C2; B3-B2))+180-180*SIGN(B3-B2)

Give this a try with several sets of locations to confirm that it’s consistently giving you the right bearing, but I believe this is the formula you’re looking for.

Gilles’ formula ATAN2 needs a comma rather than a semicolon between C2;B3
Try this
=DEGREES(ATAN2(C3-C2, B3-B2))+180-180*SIGN(B3-B2)

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Atan2 is probably a better choice since it ignores the error when divided by zero.

For the information of others
ATAN2(a,b) is equivalent to ATAN(b/a), except that a can equal 0 in ATAN2.

Steve

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Unbelievable you guys.
Steve: your explanation was extensive – Did it make it clearer for me? No, I said that it was out of my depth but your explanation sure would have assisted others.
Tim: indeed if G4 is changed to =B5/C5 (in example three) the answer is indeed correct. Thank you for pointing this out, and my apologies to Steve for not picking this issue up earlier.
Gilles: Lots of text which I will try to absorb along with the given links. Most of the text (much like Steve’s) is above me – I realise my limitations at my age.
Tim: You were quite correct in pointing out the small but significant error in Gilles’ formula, in that the comma is very necessary.

Ok, both formula appear to work and I will give it a thorough workout before I “publish” this to our members.
Thank you Steve for your formula which is located in the attached spreadsheet at A10 on all four work sheets.
Thank you Gilles for your formula which is located in the attached spreadsheet at A12 on all four work sheets.
Thank you Tim for picking up the small, but of course significant, issues that make both formulae work.

For those of you whom are interested: the UTM co-ordinates of location B in Example one (distance between UTM) is our place on planet earth. Use Google Maps/Earth to locate same (zone 56J).
Tim, you may like to read my explanation re Co-oridinates at: http://docs.google.com/View?docid=dd9r8qj3_254gfrbj4fn
Bruno Terlingen

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Hello Bruno – Thank you very much for your kind comments.I am glad that all worked out for you. I am going to read that Google Docs explanation of Co-ordinates. You guys are all way over my head on this topic.

Bruno, did you mean to attach a new Excel file to your last message?

Tim

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Perhaps this will help. I added a chart which has Loc A at the center, the NESW lines, the total distance, the Eastings/Northings differences plotted as well as the bearing angle calculated.

This should allow you to play with the values, and see the chart to see if they match what you expect. The chart is just added to what you wanted to keep. The chart data is on a separate sheet…

Steve

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My apologies for not replying sooner Tim: yes you are correct in that I somehow forgot to attach the workbook – this will be rectified with this post.

Thank you Steve, you are a gem (as per normal). I love the chart, a visual representation to our older members in the brigade will do wonders for their understanding.

I have one request Steve: The attached workbook has a lot of other data from another source, data that I have incorporated in the first worksheet. Now I don’t want to stuff up your work by trying to copy and paste same into my first work sheet (distances between UTM), so I would like to ask you to copy and paste your own two (2) worksheets into the attached workbook. I would ask you to leave range A1:J7 intact please.

I look forwards to the new workbook with your incorporated chart.

Now I owe you at least two beers when you come to OZ.
Bruno

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Dear Steve, Tim, and Gilles

I have modified the spreadsheet and incorporated Steve’s two (2) work sheets. I think that I have copied all relevant details. Would you please check that all the information/formulae are correct Steve.

I have tested this “model” with numerous known co-ordinates and the information obtained is the same as generated by the GPS device. In terms of GOTO the distances and the angles are spot on.

However (as noted on the first worksheet) this model only works when both co-ordinates are in the same UTM zone. Therefore this may need further work – anyone up to the challenge?

Please see attached file, Bruno.

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Do you have some examples and correct answers for some coordinates in other zones?

I hadn’t worked with UTM before your post and have no idea about how they are set up. The eastings and Northings in the examples here are akin to X and Y coordinates so it was easy to apply what I had learned decades ago in High School.

Steve

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Dear Steve

Please see attached spreadsheet with a number of co-ordinates in different zones, along with valid distances and bearings.
I have also made some notes within that worksheet for you to consider re learning about UTM.

Bruno

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Based on the info in the first reference:

The grid letter is unneeded since the Northings is an absolute value from the south pole so can be treated as a Y-value. It does not matter how far the grid is from the southpole as the value indicates the absolute difference from a reference point.

The grid number is important. The eastings is the distance from the left edge of the grid. So the question is, given the grid number how do you determine the distance from some arbitrary reference value to calculate to add to the eastings to get some type of X-value?

If you use the projection listed and all the grids have the same width, it is just the (grid number-1) x some width + Eastings to have it east of the left part of the map. What is the width of the grids in km? Or am I missing somethint and this is too simplistic?

If that is too simplistic, how does one calculate give the grid values for Points A & B and their eastings, what the easterly difference of the 2 points is?

Steve

PS. I did a little research and while I still don’t understand enough to do the calcs, I see my simplistic model will not work. The zones are not rectangles and the widths of the rectangles change as the Northing changes.

Also, your description in the simple explanation is not consistent with the other information I have read. The Eastings is not from the left edge of the zone but is related to the Central meridian of the Zone. And the Northing is relative to the equator, not the south pole.

Thus while you describe the coords in Zone 56J: 0397826 East & 6854848 North to be “196 Km from 150th degree East parallel of longitude” and “6854 Km and 848 metres from the south pole.” a more precise description would seem to me to be (500,000 – 397,826) = 102km 174m West of the 153rd Degree East longitude line and a distance of (10,000,000 – 6,854,848) = 3,145km 152m South of the Equator.

The across zone calcs will need to have some lookup/calculation value of the distance between the central meridians of each zone at every Northings value…

PPS
I have playing with UTMs across zones and believe it is NOT the way to go. I did find this at http://www.terrainmap.com/rm1.html which seems to me to verify my suspicions:

Although the UTM system is widely used by the USGS and others it has an inherent limitation: it is not possible to combine maps across two UTM zones without creating excessive distortion. This is because each UTM zone describes a completely different projection (remember the cylinder thing). That is, the maps are each described adequately in their own coordinate systems. It is obviously not possible to use both systems because adjacent UTM zones will each have a point described as 161,000, 3,000,000 for example but they will be different points geographically. One way to rationalize the two projections would be to choose either of the two projections and then reproject the other map onto the chosen UTM projection. For the geometric reasons mentioned earlier however, this would cause the reprojected map to be exceedingly distorted. The correct solution in this case is to choose a completely different projection that describes the entire area of interest more acceptably, and then reproject both maps onto the common optimal reference surface.

It would seem if you are going to go across zones, that you need an alternate system…

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Steve, I will respond to your text within:

Based on the info in the first reference:
The grid letter is unneeded since the Northings is an absolute value from the south pole so can be treated as a Y-value.
Bruno: True the Northings are not a problem.
It does not matter how far the grid is from the southpole as the value indicates the absolute difference from a reference point.

The grid number is important. The eastings is the distance from the left edge of the grid. So the question is, given the grid number how do you determine the distance from some arbitrary reference value to calculate to add to the eastings to get some type of X-value?
Bruno: This is indeed beyond my scope, that is why I opened the challenge.

If you use the projection listed and all the grids have the same width, it is just the (grid number-1) x some width + Eastings to have it east of the left part of the map. What is the width of the grids in km? Or am I missing somethint and this is too simplistic?
Bruno: you have come across the “cylindrical aspect”, Mr Mercator.

If that is too simplistic, how does one calculate give the grid values for Points A & B and their eastings, what the easterly difference of the 2 points is?

Steve

PS. I did a little research and while I still don’t understand enough to do the calcs, I see my simplistic model will not work. The zones are not rectangles and the widths of the rectangles change as the Northing changes.

It would seem if you are going to go across zones, that you need an alternate system…

Bruno: All true and correct.

Also, your description in the simple explanation is not consistent with the other information I have read. The Eastings is not from the left edge of the zone but is related to the Central meridian of the Zone. And the Northing is relative to the equator, not the south pole.
Bruno: True, my explanation is for a “unsophisticated” audience. I quote: “This means that I live 196 Km from 150th degree East parallel of longitude (the calculations are a bit complex here – for those who wish to know, ask me).” and indeed Steve, if one divides 0397826 by 2, one derives at a value of 196 Km – I thus took it from a central meridian as you rightly pointed out one should do. The answer in fact is spot on.

Thus while you describe the coords in Zone 56J: 0397826 East & 6854848 North to be “196 Km from 150th degree East parallel of longitude” and “6854 Km and 848 metres from the south pole.” a more precise description would seem to me to be (500,000 – 397,826) = 102km 174m West of the 153rd Degree East longitude line
Bruno: You could have a point here, I only looked at the 150th degree East parallel of longitude – see my explanation above.
and a distance of (10,000,000 – 6,854,848) = 3,145km 152m South of the Equator.
Bruno: I live at -28.42 degrees lat, therefore 28.42*111 (111 Km per degree lat) = 3152.62 Km – which is close enough to your calculations.
The across zone calcs will need to have some lookup/calculation value of the distance between the central meridians of each zone at every Northings value…

PPS
I have playing with UTMs across zones and believe it is NOT the way to go. I did find this at 1. http://www.terrainmap.com/rm1.html which seems to me to verify my suspicions:

Bruno: Well Steve, it was a challenge. Like I mentioned on my site, we will in all likelihood only ever fight fires in zone 56J.
It would seem if you are going to go across zones, that you need an alternate system…

Bruno: Any suggestions?

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I suggest converting the UTM to latititude and longitude and then use spherical calculations

Some examples are in http://www.movable-type.co.uk/scripts/latlong.html even with Excel formulas.

If you are using these large distances, I think you should throw away your simple planar representations/approximations. You are trying to look at the 2 points as the diagonal of a square, when a better planar approximation is the diagonal of a trapezoid. The distance North and South can be determined directly, but the distance East and West is different (If Point B is NE direction from A, it is shorter to travel East from A until due South of B then go N, then it would be to travel Due N until you are due West of B and then go East. The trapezoid is bigger closer to the equator.

If you want to stick to planar representations, use the trapezoid, though if you are going to use excel I see no reason not to take the next step and presume a sphere since you will be using Lat and long anyway. The final correction will have to be adjustment for magnetic vs true north. The calcs in the page are for true north, but compasses don’t read true north…

Steve

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Dear Steve
I have several workbooks with lots of GPS data, one worksheet deals with the Great Circle Distance. Like you probably, I learned this YEARS ago and now I have a simple fill in re Lat/Long fields that automatically calculates the distance, but of course not the Bearing.

OK, I am now working on a Sun Dial for my co-ordinates. I have lived here only for the past two years. I made out a spreadsheet which determined the scribe angles for specific times of the daylight hours about five years ago (for a different latitude) and I am having trouble working out what I did back then. If I can’t work it out I will commence a new thread in this forum – may see you then.

Bruno.

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Bruno:

I agree with Steve. If you’re scope is larger than a few miles, you should use Great Circle formulas. OK, they’re a little more complex than right triangles. And, they’re not as accurate as Geodesic formulas. (Because the Earth isn’t a perfect sphere. I think they call it an oblate elliptoid.) But, they’re (relatively) short and tidy formulas, and every conceivable variation on a theme has been worked out. As a pilot, I had an interest in navigation since the early 1970s, and have transposed the spherical triginometry formulas (the same ones used in celestial navigation) to get exactly what I needed for my applications.

Who cares how complicated the formulas are, if they work, anyhow? Once you go through the effort to program your spreadsheet and make it bulletproof, you’ll be set forever. You wouldn’t have to reinvent the wheel. All the forumulas are out there floating around the Internet right now.

And a big advantage is the latitude/longitude coordinate system is in widespread use. The comprehension gap for the end user would, I think, be narrower than the UTM system. The way I understand it, the whole intent of the UTM system was to simplify calculation of bearings and distances using the Pythagorean Theorem. Now, with computers, this original rationale is less important. Since you’re using Excel to do the math, the way I see it, the sky’s the limit. No pun intended. Think big! You’ve got some real brainiacs on this forum to tap into. (Not referring to myself, of course :))

Steve’s point about the magnetic variation is a good one. What method would the firefighters use to navigate to the target location? Would they be using a compass? If so, how would they measure distance? Or would they have GPS receivers?

I recall some correspondence I had with a theoretical physicist at the Lawrence Livermore Laboratories who had taken a database of the world’s magnetic variation and put it into a monster formula. One could then use the formula to calculate magnetic variation for any point on the globe (within a certain tolerance, of course).

Magnetic variation has its own problems, of course. It changes with time. We’re in the process of flipping North and South poles. Of course, it won’t happen for a few more years yet. I think it happened in the movie 2012, which gives us a couple of years.

But, I digress. It’s going on 3AM here, and I’m rambling. Excuse me!

Sounds like a fun project. Good luck.

Bill

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Thank you Bill for your thoughts.
I agree with most of your observations. Let me tell you the story: our fire brigade has been issued with two (2) Garmin eTrex (H) model GPS devices. I have used these devices since 1998 and am conversant with both lat/lon and UTM. Most city map here use UTM only while new country maps have a tendency to show both lately. I know that with the personnel involved there is going to be confusion as to send and receive information re co-ordinates to the authorities (whom themselves are not up-to-date). The eTex is not a compass yet can be used to determine the bearing of a fire in case the personnel need to do a “projection”. To this end I am writing a manual for our brigade members so that they can use the eTrex with some degree of confidence. I can’t expect “country boys” to take a laptop to the fire, open a U-beaut spreadsheet, fill in fields and then determine Great Circle Distances. What I can show them, using UTM measurements, is basic arithmetic to determine a distance on paper – however none of this is needed as the eTrex will show all this information to the user.

The last link that Steve provided gave some interesting information, in that the bearing actually changes over a great distance – something that had never crossed my mind. I quote: “In general, your current bearing will vary as you follow a great circle path (orthodrome); the final bearing will differ from the initial bearing by varying degrees according to distance and latitude (if you were to go from say 35°N,45°E (Baghdad) to 35°N,135°E (Osaka), you would start on a bearing of 60° and end up on a bearing of 120°!).”

No Bill, I do not want to reinvent the wheel, I would just like to keep the tyre pressure up! (tyre in Oz and tire in USA).

Bruno.

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