# Locating an Object in 3-D Space.

Discussion in 'Math' started by SplitInfinity, Jun 2, 2013.

1. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
OK...let's see who knows this here.

We have a glass box. And in this box we have an INVISIBLE OBJECT.

Now the object here is to locate this invisible object with Geometry and a few items.

The Glass Box is in space and no Gravity issues arise. As well the invisible object is static within this box.

On the inside surface of the box are 6 hooks and each hook is located at a position that is known and relative to the Invisible Object.

We have 6 strings.

Now using the LEAST number of strings possible and the least number of Hooks/Known Relative Points of Position to the Invisible Object....FIND THE LOCATION OF THE INVISIBLE OBJECT.

Detail how you did it and how many strings and hooks used.

Split Infinity....p.s...Bonus Points....detail the total amount of Points of Position required to locate the Invisible Object and Chart a Course to it.

2. ### LDC3 Active Member

Apr 27, 2013
920
160
All that is needed is 4 non-collinear, non-coplaner points (hooks) and 4 strings to determine the position of the object. The first position defines a sphere of points where the object can be. The second point (along with the first) defines an circle of points where the object can be. The 3rd point (since it is non-collinear) forces the object to be in one of 2 places. The 4th point restricts the location to one point.

3. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
OK...you are not capable of doing what you state here with what you have been given.

All you have is that Glass Box...6 known points of position and known as to their relative positions are to the invisible object...the hooks and the strings.

Split Infinity

4. ### LDC3 Active Member

Apr 27, 2013
920
160
I beg to differ. The distance from the first point (hook) define a sphere of locations for the object. The distance from the second point (another sphere) intersects the first sphere and defines a circle where the 2 spheres intersect. The distance from the 3rd point (another sphere) intersects the circle in 2 locations. The distance from the 4th point (yet another sphere) only intersects one of the points since the 4 starting points are non-collinear and non-coplaner.

5. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
I will give a hint...you will not need to use a few of the strings...LOL!

Split Infinity

6. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
Sorry...this will not work.

Remember you have no idea where this invisible object is but you DO have 6 points of position both known to you and known and relative to the invisible object.

Split Infinity

7. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
Exra Hint...it does not really matter that the object is in a box as it could be in any geometric 3-D shape that has volume as well the hooks could be static as well and not attached to the box or there does not need to be a box at all.

Split Infinity

8. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
I should add that the six points of position are around the object.

Split Infinity

9. ### djsfantasi AAC Fanatic!

Apr 11, 2010
2,911
881

What do you mean by "and known as to their relative positions are to the invisible object"?

And what do you use the strings for? Are they lines calculated between hooks and/or the invisible object?

I think you can do it with two hooks and somewhere between 2 and 4 strings, dependent on how you define them.

Two hooks and the invisible object define a plane, simplifying the problem. Then, with two known points and a third unknown, it is possible to calculate the position of the third. Converting back from the defined 2D plane to 3D will give you the answer.

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10. ### amilton542 Active Member

Nov 13, 2010
494
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What's your definition of object location? Are you referring to the centre of this object relative to some Origin?

Personally, there's an infinite number of solutions to this problem. It's like me saying, "I've got six wall-outlets in my bedroom, where did I leave my car keys?"

mikeleeson, shortbus and studiot like this.
11. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
OK...since some are getting confused by the terminology...or perhaps I did not do a good enough job describing it...I will just come out and explain it.

In order to find a position or object or celestial body in 3-D space...from any other location or distance in any direction surrounding that object position...you need 6 known and relative points of position to the object.

So say I was in a spacecraft in a different sector of space than Earths...and I was trying to travel to a star system who's star was obscured from detection from my location by say the Horse Head Nebula in Orion.

Now I already know 6 stars that are of sufficient magnitude to be visible from any direction or location from this hidden star and these 6 stars are my 6 relative points of position that are of known distances and vectors from this hidden star so that by simply drawing a straight line from one star to another and doing this 3 times for three sets of stars....the three straight lines will cross and at this point will be located the hidden star system which I am trying to travel to....a 7th point of position would be from where I am currently at in the craft and by drawing a straight line from that point to the cross points of all three lines I will have plotted a course.

Split Infinity

Mar 3, 2013
369
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13. ### LDC3 Active Member

Apr 27, 2013
920
160
It is very rare that 3 stars line up in a collinear fashion. And even rarer for 3 lines between 6 stars to intersect at one location. Even if the location is expanded to a 1 light year radius sphere.

You may wish to navigate like this, but in reality, it cannot be done.

14. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
1,305
Detail how I did it;
1. place three hooks in optimum locations in the box corners (on 1 face).
2. measure the length of the 3 strings to the object.
3. triangulate.

The LEAST number of strings possible is 3, and it is POSSIBLE because the object is constrained within the confines of the box.

15. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
The stars used as the 6 points of position will be multiple if not over a hundred light years distant.

Whether or not one would use this for navigation matters not...if you were lost in space...and needed to get somewhere...and had no other method of finding out where the star system is you needed to get to...have the ability to just be able to find bright magnitude stars that are known and relative to the place you needed to go...and by using some simple geometry be able to plot a course there...well I for one would definitely want to know how to do this.

Split Infinity

16. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
9
RB...perhaps I was not clear.

The 6 hooks are attached to the insides of the glass box at 6 points of position which are positions known and relative as to their distance and vectors away from the invisible object.

By connecting a string between two hooks and doing this twice more...so three strings each one connected to two hooks and doing so in a manner where all three strings cross one another...at the point of their crossing will be the location of the invisible object.

See the diagram I posted.

Split Infinity

17. ### THE_RB AAC Fanatic!

Feb 11, 2008
5,435
1,305
Thanks for the correction. I assumed the "strings" were distance measuring strings as you mentioned "geometry", and nothing in your post said otherwise.

18. ### djsfantasi AAC Fanatic!

Apr 11, 2010
2,911
881
Sorry, I am still not clear. In my attached picture, this is how I see the problem. There is no obvious correlation between colinear stars and the location of the unknown object. Am I missing a constraint from the original problem?

Note that in my example, there is no intersection of the string from the front face to the back face and the string from the left face to the right face...

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19. ### SplitInfinity Thread Starter Member

Mar 3, 2013
369
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The reason why you are not getting this is probably my fault as I should have explained the term...Known and Relative...as it applies to the 6 hooks or positions.

Known and Relative is describing the positions of these 6 hooks or 6 points of positions around the invisible object. I will use this example. If it is impossible using any other means to find the location of an object in any 3-D space...and the object within that space does not have to be inside the bounderies may they be physical or otherwise...of any geometric shape with a volume...then as long as you know 6 visible or detectable points of position that are known and relative....ie....known...the distance between these 6 points of position or objects and the invisible object....relative....the angles or vectors of each 6 objects or points of position to the invisible or undetectable object and to each other...just by drawing 3 lines...each line connecting two points of the 6 points and where all three of these lines cross will be the invisible object.

You cannot just use ANY 6 points or objects around the hidden or invisible object to do this...these 6 points would at first be CHOSEN as a way to allow this invisible object to be found.

This method is used when Triangulation is not possible or a radio signal or line of site or radar is not possible.

If I was in space and the star system I needed to get to was obscured by a Nebula or a Star Cluster...but what has already been done is for every Celestial Body of importance 6 points of position...those points being such objects as Bright Magnitude Stars...Bright Distant Galaxies...etc...have been used to provide a method of location for either very dim stars or other celestial bodies or when such bodies might be obscured from either detection or view due to Gas Nebula, Particle Debris or perhaps they might be behind a very bright star cluster of near star.

So using simple geometry no matter from what angle or location or even great distance as such 6 points are usually chosen so that no matter where one might be in the Galaxy...and this works when Deep Space Probes need to self align to send and receive signals from Earth and although Pulsar based calculations can be done this assumes that such a probe can still receive signal even if it can send one. If it can send but not receive...and such a probe as the newer ones are to a point self-determinate...by using such a 6 point system no matter where the probe maybe from Earth and no matter at what vector or angle....by the probe looking at star positions...using the 6 point calculation it will ALWAYS know where to point and send a signal even if it cannot receive a signal.

Hope that helped.

Split Infinity

20. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
515
I have been watching the development of this thread with interest.

There is, of course, a vital missing piece of information.

The size of the 'object' in the glass box. It appears to have miraculously shrunk to a point in all analyses so far presented.

If the distance and directions are known to six known points in 3D space the information (for a point) is redundant and surveyors would employ some statistics to complete the task.