Solving for tent coordinates

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Main Question or Discussion Point

About 30 years ago (long before GPS!) we pitched our tent in a mountainous wilderness area. I took a photograph of the mountain with our tent in the foreground (attached). We now, for nostalgic reasons, would like to pitch our tent again on exactly the same spot, or at least as close to it as possible. I have a rough idea where the spot is to within several hundreds of meters, but I would like to do better than that. Could you perhaps please assist with that?

There are a number of well known peaks on the photo whose GPS coordinates and height above sea level are known. It should be therefore possible that the various angles that I could draw on a copy of the photo with lines from peak to tent to another peak would provide enough information to set up a number of equations to solve for the two unknowns i.e. the x and y coordinates of the tent position.

One assumption that I have to make is an estimate of the height of the tent above sea level in order to get the difference in height of the various peaks above that of the plane of the tent. That should not introduce a major uncertainty since I have a good topographic map with 20 meter contours and the terrain is reasonably flat over the broader area of where the site is likely to be.

One can then construct a number of right angled triangles with each perpendicular triangle passing through the tent and a peak (see figure). The one side of a given triangle is the height of the tip of the peak, the hypotenuse is the line from the peak tip to the tent. The third side is the line from the tent to the base of the peak as projected onto the horizontal plane of the tent. The lengths of some of the sides can be expressed in terms of the unknown tent coordinates.

The angles that are measurable from the photo are the angles between the various hypotenuses. Application of the cosine rule should then relate the value of the cosines of the angles on the photo to the lengths of the sides of the relevant triangles that contain the unknown coordinates.

Unfortunately I find that my equations are quadratic terms divide by the square root of the product of quadratic terms, which I cannot solve analytically. That would call for numerical methods.

Furthermore, an error is probably introduced in my approach because the assumption is that the angles as measured on the photo are the true angles as if the camera was in a position directly above the tent pole. In fact, I was standing ten or more meters away from the tent. Hopefully that is a small distance as compared to the thousands of meters to the peak tips. That then may or may not influence the size of the measured angles thus introducing an error.

Do you perhaps see a different way of how to tackle the problem of solving for the unknown coordinates so that I could enter the solution into my GPS and walk straight to the spot? It should be a piece of cake for people familiar with the maths of the far more complex situation of how GPS satellite info is derived.

To get around that uncertainty one might rather draw the lines towards the actual camera position on the photo and then place the tent about 10 meters from there after solving for the camera position. However, where on a photo is the point of the position of the camera? Presumably it is off the photo, but where? I could have taken an identical photo more than a kilometre away if I had a very long telephoto lens. Where would you then place the camera on the photo relative to the tent? It seems that the focal length of the camera must somehow then enter into the discussion as well.

Thank you for your attention.

Regards.
Wim
 

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  • #2
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This is interesting! Can you post known coordinates (including altitudes) for the peaks you've labeled?

I think for the reasons you gave in your last paragraph, there will be considerable uncertainty along the direction the camera faces. But if there's enough variation in distance from the camera among the visible peaks, it should be possible to get some information along that direction, too.
 
  • #3
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Thanks for your interest Tinyboss.

I shall post the details shortly. I need to verify the exact position of the left peak on the photo. The peak's highest point is not visible on the horizon since it is somewhat back from the edge. The line as I drew it on the photo might therefore not be fully correct.
 
  • #4
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Here are the relevant quantities.

The peaks that I have data for are from left to right CC, MC,CP,SH, Tu and Am. Ignore the two lines between MC and CP.

ANGLES: The values of the angles as I measured them on the photo (peak1 to tent to peak2) are:
CC-MC: 6.9 deg. MC-CP: 22.0 deg. CP-SH: 40.5 deg. SH-Tu: 40.1 deg. Tu-Am: 3.9deg.

The difference in the angles as measured with a protractor compared to measuring the lengths of the lines, plus the connecting line, and then calculating the angle via the cosine rule, was at maximum 0.5 degrees. The position of CC on the photo is somewhat uncertain as it is not seen on the photo since it is positioned somewhat back from the edge and I made a reasonable guess.

The peak heights above sea level as stated on a topographic map are:
CC - 3246m MC - 3229m CP - 3148m SH - 2973m Tu - 2670m Am - 2620m

My estimate of the height of the tent position is 2164m, which is probably within 20m or so. What is of relevance for calculations would be the differences between the various peak heights and of the tent (2164m) and also between themselves.

I derived the coordinates of the peaks in degrees by interpolation from a map that was drafted based on the so-called CAPE DATUM. The peak positions would therefore be in error if they were entered into a GPS that is set to the WGS84 DATUM. The difference apparently might be up to 100m in that area. This is not a problem since a solution in one DATUM can always be converted back.

What I have done, though, was to select an arbitrary zero position on the map and measured in meters the various indicated peak base positions from the arbitrary origin and also the distances between the bases of the peaks. That origin was chosen as: 29 deg 4.0 minutes South and 29 deg 21.0 minutes East. These distances can be used in the calculations. Once the coordinates of the tent position from the arbitrary origin are solved one can convert back to degrees.

I list the map derived coordinates (in degrees) below in case you are interested just to take a quick look at the area in Google Earth. The peaks will then be something like 100m or so to the left of where they will be indicated in GE:

CC : 29 20.762’E 29 5.257’S MC: 29 20.747’E 29 4.791S
CP: 29 21.115’E 29 4.490’S
SH: 29 21.123’E 29 4.061’S Tu: 29 21.284’E 29 3.679’S
Am: 29 21.376’E 29 3.365’S

The coordinates of the peaks in terms of meters from the convenient arbitrary origin are:
CC: (-383, -476) MC: (-409,371) CP: (192,933) SH: (204, 1724) Tu: (452,2419)
Am : (600,2997). The negative signs indicate that the peak lies to the West and South the origin.

Horizontal distances between peaks (d31, d12, etc):
They were measured on the 1:50 000 map.
CC - MC: 850m MC- CP: 814m CP – SH: 792m SH – TU: 753m Tu – Am: 586m

I calculated the direct distances between the peak tops from the difference in heights of the two peaks plus also their horizontal distance from one another. For two peaks that do not differ much in height the value will not differ much from the distance between the bases of the peak. The derived values are:
CC- MC: 850m MC – CP: 818m CP – SH: 811m SH – Tu: 812m Tu – Am: 588m.

Calculation of angles alpha:
In a triangle such as Tent – B – C, for example, the angle alpha12 can be expressed in terms of its three sides via the cosine rule. The one side of the triangle, e.g. BC , is known. The other two sides of the triangle can be expressed in terms of differences involving the unknown coordinates (xt,yt) of the tent. Two equations are required to solve for the two unknowns. However, several such equations can be drawn up since there are six peaks thus five angles can be measured thus giving five equations and two unknowns.
 

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