File:Trilateration picture.png
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Revision as of 19:09, 16 December 2010 by JoeEisner (talk | contribs) (uploaded a new version of "File:Trilateration picture.png": Reverted to version as of 19:07, 16 December 2010)
Size of this preview: 800 × 572 pixels. Other resolutions: 320 × 229 pixels | 1,394 × 996 pixels.
Original file (1,394 × 996 pixels, file size: 58 KB, MIME type: image/png)
The target to be located lies directly on the intersection of the circles. In order to make this calculation in 2D space, we need three of the red points. For instance, if we only had two of the red points, there would be two intersections, A and A’, and with out other constraints (such as the target must be below the dotted line that goes through points 1 and 2) the location could not be determined.
File history
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| Date/Time | Thumbnail | Dimensions | User | Comment | |
|---|---|---|---|---|---|
| current | 19:10, 16 December 2010 | | 1,394 × 996 (58 KB) | JoeEisner (talk | contribs) | Reverted to version as of 19:08, 16 December 2010 |
| 19:10, 16 December 2010 |  | 1,394 × 996 (58 KB) | JoeEisner (talk | contribs) | Reverted to version as of 19:07, 16 December 2010 | |
| 19:09, 16 December 2010 |  | 1,394 × 996 (15 KB) | JoeEisner (talk | contribs) | Reverted to version as of 19:06, 16 December 2010 | |
| 19:09, 16 December 2010 |  | 1,394 × 996 (58 KB) | JoeEisner (talk | contribs) | Reverted to version as of 19:07, 16 December 2010 | |
| 19:08, 16 December 2010 |  | 1,394 × 996 (58 KB) | JoeEisner (talk | contribs) | The target to be located lies directly on the intersection of the circles. In order to make this calculation in 2D space, we need three of the red points. For instance, if we only had two of the red points, there would be two intersections, A and A’, an | |
| 19:07, 16 December 2010 |  | 1,394 × 996 (58 KB) | JoeEisner (talk | contribs) | The target to be located lies directly on the intersection of the circles. In order to make this calculation in 2D space, we need three of the red points. For instance, if we only had two of the red points, there would be two intersections, A and A’, an | |
| 19:06, 16 December 2010 |  | 1,394 × 996 (15 KB) | JoeEisner (talk | contribs) | The target to be located lies directly on the intersection of the circles. In order to make this calculation in 2D space, we need three of the red points. For instance, if we only had two of the red points, there would be two intersections, A and A’, an |
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