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Where on Earth II (Posted on 2005-01-16) Difficulty: 3 of 5
Is it possible to determine the location of an object, if:-

The date is 21st July
The object is 5 Metres tall
The Shadow at 9AM GMT is 31.75 metres long,
The Shadow at Noon GMT is 6.22 metres long,
The Shadow at 3PM GMT is 2.68 metres long.

(On 21 July, the sun is 20.4 north of the equator and crosses the meridian 6.3 minutes after local noon.)

No Solution Yet Submitted by Juggler    
Rating: 4.0000 (1 votes)

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Solution solution Comment 4 of 4 |

It looks like most everyone has sworn off spherical trig since Location X.

When the object's shadow is 31.75 m long, the sun is arctan(5/31.75) degrees up in the sky. More pertinantly the location is the complement of this, or arctan(31.75/5) = 81.0505 degrees of great circle arc away from the point where the sun is directly overhead--called the subsolar point.  Tabulating these arc distances:

9 AM GMT: 81.0505
Noon GMT: 51.2056
3 PM GMT: 28.1913

We are taking the latitude of the subsolar point as being 20.4 degrees north all day, placing it 69.6 degrees from the north pole.  At 9 AM GMT (the time zone for the prime meridian, 0 degrees longitude), the subsolar point's longitude would be 45 degrees east if the sun crossed the meridian at local noon, but on this date, the sun passes any meridian 6.3 minutes after local noon.  As one day (1440 minutes) corresponds to 360 degrees, 6.3 minutes represents 1.575 degrees, so the subsolar point at 9 AM GMT is at 46.575 degrees east (it has not yet crossed the 45th east meridian).

We want to determine the location of the object relative to this meridian and then we can adjust to the standard, Greenwich meridian knowing the longitude of the subsolar point.  We can see that the sun is getting higher in the sky even after Greenwich noon, so the location must be west of Greenwich, so that its time zone places 3 PM GMT closer to local noon, so whatever relative longitude west of the first subsolar point we find, we'll subtract 46.575 degrees to get the standard western longitude relative to Greenwich.

We need to do some spherical trigonometry.  Our first spherical triangle will have vertices at the north pole, P; the subsolar point at 9 AM GMT, S1; and the subsolar point at Noon.  As these points will also be used in other triangles later, we'll use different letters to refer to the measure of the angles at these points as part of the triangle. The measure of the angle at the pole we'll designate A; at S1, we'll designate B; and at S2, C. We'll follow the usual convention of labeling the opposite sides with the lower case of the angle opposite.

As the sun travels 45 degrees in 3 hours, A = 45 degrees. The sides b and c are the distance of the subsolar points from the pole: 69.6 degrees each.  The spherical law of cosines can be used to find side a: cos a = cos b cos c + sin b sin c cos A. Side a comes out to  42.0385858 degrees (extra precision carried on intermediate results, to be rounded later).

Then angle B can be calculated from cos b = cos a cos c + sin a sin c cos B. It comes out to 81.7842242 degrees.

Then we'll introduce point X, the unknown location of the object.  As we know the angular arc distance from each of the two subsolar points S1 and S2, we'll make X, S1 and S2 the vertices of another triangle.  We'll call the measures of these angles X, D and E respectively.  Note that D is at the same vertex as C, but a different measure as it goes to the unknown object rather than to the north pole. E is at the same vertex as B but again for the same reason is a different measure.

Side d has measure 81.0505 degrees, from the table of zenith distances above. Side e has measure 51.2056 degrees. Side x is the same as side a in the first triangle: the distance from S1 to S2, 42.0385858 degrees. Using cos e = cos d cos x + sin d sin x cos E, we can find E. It's 39.4212206 degrees.

What we are doing is finding the intersection of two circles--one centered on subsolar point S1, and the other centered on subsolar point S2.  In general, two circles intersect at two points.  The purpose of the third reading (the 3 PM GMT reading) is to differentiate between these.  Right now, in our calculation, we have to decide (actually we have to do it both ways, but one at a time) which to calculate.  The line from S1 to location X might pass above (north of) the line from S1 to S2, or below (south of) that line. 

The next (and last) triangle we are to consider is that whose vertices are P (the north pole), S1 and X. We'll call the measur of the angle, in this triangle, at the pole to be H; the angle at S1, F; and the angle at location X, G.  We had two previous angles' measures at S1: B and E.  Now, depending on whether the line from S1 to S2 goes north of the line from S1 to S2 or south of that line, angle F could be either B - E or B + E, respectively. Each will give a location of the intersection of the two circles; then we'll have to check each one for consistency with the 3 PM GMT reading.

Side h is the same 81.0505 degrees as it is coincident with side d in the previous triangle.  Side g is the same as side c in the first triangle, 69.6 degrees. 

Taking the first scenario of F = B - E = 42.3630036, f, the distance of location X from the north pole, can be calculated by cos f = cos g cos h + sin g sin h cos F. It comes out to f = 42.4095572, corresponding to a latitude of 47.5904428 (its complement), or rounded to 47.59 degrees north.

Angle H, again using the law of cosines, cos h = cos f cos g + sin f sin g cos H, comes out to 99.2676428.  If we subtract 46.575 from this, as we said we would to make the longitude relative to Greenwich, that comes out to 52.6926428 degrees west, rounded to 52.69.

The second scenario would have F = B + E = 121.2054448 degrees.  Using the same formulae, f comes out to 115.1808228, corresponding to 25.18 degrees south, and H comes out to 69.007149, corresponding to a longitude of 22.43 degrees west.

To differentiate between these we need to find the zenith distance of the sun at 3 PM GMT at each of them. Since H is the difference in longitude between S1 and X, we just need to subtract 90 degrees from each version's H, and recalculate h, using the law of cosines, to match against the expected zenith distance.

For 47.59 N, 52.69 W, we get 28.2075, quite close to the zenith distance we expect at 3 PM GMT. At 25.18 S, 22.43 W, we get a zenith distance of 49.9390 degrees.  So the answer is 47.59 N, 52.69 W.  Looking on a map, this seems to be around St. John's, Newfoundland, Canada.

The calculations were actually done by a triangle solving program.  They are shown below. Note the negative H and h in the final, alternative, location calculation, representing only that the sun had already passed the meridian (it was afternoon) at that location at 3 PM GMT.

SPH:Command :tri:abc
SPH:Command :A=45
SPH:Command :b=69.6
SPH:Command :c=b
 69.59999999999999
SPH:Command :sol:bAc
A= 45 B= 81.78422421314821 C= 81.78422421314821
a= 42.03858581089508 b= 69.59999999999999 c= 69.59999999999999
SPH:Command :x=a
 42.03858581089508
SPH:Command :d=81.0505
SPH:Command :e=51.2056
SPH:Command :tri:xde
SPH:Command :sol:xde
D= 126.4059105654866 E= 39.42122059900704 X= 33.06449892498474
d= 81.0505 e= 51.2056 x= 42.03858581089508
SPH:Command :
INVALID COMMAND!
SPH:Command :F=B-E
 42.36300361414116
SPH:Command :h=d
 81.0505
SPH:Command :g=c
 69.59999999999999
SPH:Command :tri:fgh
SPH:Command :sol:hFg
F= 42.36300361414116 G= 69.46334495309584 H= 99.26764280927183
f= 42.40955716459464 g= 69.59999999999999 h= 81.0505
SPH:Command :L=H-45-1.575
 52.69264280927183
SPH:Command :l=90-f
 47.59044283540536
 
SPH:Command :H=H-90
 9.267642809271834
SPH:Command :sol:Hfg
F= 13.28468449083447 G= 161.3763936801385 H= 9.267642809271834
f= 42.40955716459464 g= 69.59999999999999 h= 28.20752978152254
SPH:Command :
INVALID COMMAND!
SPH:Command :F=B+E
 121.2054448121552
SPH:Command :h=d
 81.0505
SPH:Command :g=c
 69.59999999999999
SPH:Command :sol:Fhg
F= 121.2054448121552 G= 62.35686816485986 H= 69.0071492267241
f= 115.1808228010735 g= 69.59999999999999 h= 81.0505
SPH:Command :H=H-90
-20.9928507732759
SPH:Command :sol:Hfg
F= 25.06209520051021 G= 153.9774317971655 H=-20.9928507732759
f= 115.1808228010735 g= 69.59999999999999 h=-49.93902303402519

Note that the solution is based on taking the midline of the edge of what in some instances is a fuzzy shadow. The sun's diameter in the sky is about 1/2 degree.  When the shadow is about 32 meters long its fuzziness is about 1 meter, so the best way of determining its end point might be to put on welders glasses and look up from the ground to see where the object obscuration bisects the sun's disk.


  Posted by Charlie on 2005-01-20 14:17:53
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