How to find out if a set of point3D are in the same plane?












2















I have a set of point3D (X, Y, Z). I need to check if they are coplanar with some sort of tolerances. My way of doing it is that: I convert all points from the Global Coordinate System to Local one, where local x,y is in the same plane of the plane defined by 3 points in the set, and z is normal to that plane. And then, all I need to do is to check if all points in the set have approximately similar local z values.



However, the tricky part is how to pick the 3 points to define the reference plane. If picked randomly, this would result in sometimes the set of points are coplanar, sometimes not. Do you have any suggestion?






algorithm geometry point plane geometry-surface







share|improve this question



























    2















    I have a set of point3D (X, Y, Z). I need to check if they are coplanar with some sort of tolerances. My way of doing it is that: I convert all points from the Global Coordinate System to Local one, where local x,y is in the same plane of the plane defined by 3 points in the set, and z is normal to that plane. And then, all I need to do is to check if all points in the set have approximately similar local z values.



    However, the tricky part is how to pick the 3 points to define the reference plane. If picked randomly, this would result in sometimes the set of points are coplanar, sometimes not. Do you have any suggestion?






    algorithm geometry point plane geometry-surface







    share|improve this question

























      2












      2








      2








      I have a set of point3D (X, Y, Z). I need to check if they are coplanar with some sort of tolerances. My way of doing it is that: I convert all points from the Global Coordinate System to Local one, where local x,y is in the same plane of the plane defined by 3 points in the set, and z is normal to that plane. And then, all I need to do is to check if all points in the set have approximately similar local z values.



      However, the tricky part is how to pick the 3 points to define the reference plane. If picked randomly, this would result in sometimes the set of points are coplanar, sometimes not. Do you have any suggestion?






      algorithm geometry point plane geometry-surface







      share|improve this question














      I have a set of point3D (X, Y, Z). I need to check if they are coplanar with some sort of tolerances. My way of doing it is that: I convert all points from the Global Coordinate System to Local one, where local x,y is in the same plane of the plane defined by 3 points in the set, and z is normal to that plane. And then, all I need to do is to check if all points in the set have approximately similar local z values.



      However, the tricky part is how to pick the 3 points to define the reference plane. If picked randomly, this would result in sometimes the set of points are coplanar, sometimes not. Do you have any suggestion?






      algorithm geometry point plane geometry-surface




      algorithm geometry point plane geometry-surface






      share|improve this question













      share|improve this question











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      asked Jan 2 at 22:29









      N.T.CN.T.C

      3412515




      3412515
























          2 Answers
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          Probably the most common way to do this would be with Principal Component Analysis: https://en.wikipedia.org/wiki/Principal_component_analysis



          The short description is:




          1. Calculate the 3x3 covariance matrix of the input points

          2. Extract the smallest Eigenvector. That is the normal vector for the plane that is the Least-Squared-Error fit to all your points.






          share|improve this answer































            0
















            1. normal n



              pick any 3 points from your dataset that are not on a single line let call them p0,p1,p2. To improve accuracy they should be more distant to each other. Now to construct a normal vector do this:



              n = cross( p1-p0 , p2-p0 ); // perpendicular vector to both operands of cross
              
n /= |n|;                   // unit vector



            2. check all points



              for any point on the plane formed by p0,p1,p2 the altitude component (in normal direction) should be zero so for any point p:



              |dot( p-p0 , n )|<=1e-10


              the 1e-10 is just some zero threshold due to accuracy loss on FPU ... Any point not satisfying the condition does not belong to the plane ...




            So how to pick 3 points forming a triangle ?





            1. pick p0,p1



              pick p0 as point that has minimal x,y,z coordinates and p1 that has maximal x,y,z coordinates. This ensures they are distant enough.




            2. pick p2



              now sear the points and find such that |dot( p1-p0, pi-p0 )| is minimal. While |pi-p0| is not zero and big enough (for example at least 0.1*|p1-p0|). That ensures the points form a triangle and are not too close to each.




            All of this is doable in O(n) so is still fast ...






            share|improve this answer


























            • Isn't this same as what I described in my question? The problem here is actually what criterion to use to pick p1, p2 and p3. It is quite tricky since, for example, if all points are co-planar but p3, then the pick would result in wrong results.

              – N.T.C
              Jan 3 at 8:53











            • @N.T.C Yes youre right sorry I forgot to submit the other part of answer ... silly me :)

              – Spektre
              Jan 3 at 8:59














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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            3














            Probably the most common way to do this would be with Principal Component Analysis: https://en.wikipedia.org/wiki/Principal_component_analysis



            The short description is:




            1. Calculate the 3x3 covariance matrix of the input points

            2. Extract the smallest Eigenvector. That is the normal vector for the plane that is the Least-Squared-Error fit to all your points.






            share|improve this answer




























              3














              Probably the most common way to do this would be with Principal Component Analysis: https://en.wikipedia.org/wiki/Principal_component_analysis



              The short description is:




              1. Calculate the 3x3 covariance matrix of the input points

              2. Extract the smallest Eigenvector. That is the normal vector for the plane that is the Least-Squared-Error fit to all your points.






              share|improve this answer


























                3












                3








                3







                Probably the most common way to do this would be with Principal Component Analysis: https://en.wikipedia.org/wiki/Principal_component_analysis



                The short description is:




                1. Calculate the 3x3 covariance matrix of the input points

                2. Extract the smallest Eigenvector. That is the normal vector for the plane that is the Least-Squared-Error fit to all your points.






                share|improve this answer













                Probably the most common way to do this would be with Principal Component Analysis: https://en.wikipedia.org/wiki/Principal_component_analysis



                The short description is:




                1. Calculate the 3x3 covariance matrix of the input points

                2. Extract the smallest Eigenvector. That is the normal vector for the plane that is the Least-Squared-Error fit to all your points.







                share|improve this answer












                share|improve this answer



                share|improve this answer










                answered Jan 2 at 23:00









                Matt TimmermansMatt Timmermans

                21.1k11733




                21.1k11733

























                    0
















                    1. normal n



                      pick any 3 points from your dataset that are not on a single line let call them p0,p1,p2. To improve accuracy they should be more distant to each other. Now to construct a normal vector do this:



                      n = cross( p1-p0 , p2-p0 ); // perpendicular vector to both operands of cross
                      
n /= |n|;                   // unit vector



                    2. check all points



                      for any point on the plane formed by p0,p1,p2 the altitude component (in normal direction) should be zero so for any point p:



                      |dot( p-p0 , n )|<=1e-10


                      the 1e-10 is just some zero threshold due to accuracy loss on FPU ... Any point not satisfying the condition does not belong to the plane ...




                    So how to pick 3 points forming a triangle ?





                    1. pick p0,p1



                      pick p0 as point that has minimal x,y,z coordinates and p1 that has maximal x,y,z coordinates. This ensures they are distant enough.




                    2. pick p2



                      now sear the points and find such that |dot( p1-p0, pi-p0 )| is minimal. While |pi-p0| is not zero and big enough (for example at least 0.1*|p1-p0|). That ensures the points form a triangle and are not too close to each.




                    All of this is doable in O(n) so is still fast ...






                    share|improve this answer


























                    • Isn't this same as what I described in my question? The problem here is actually what criterion to use to pick p1, p2 and p3. It is quite tricky since, for example, if all points are co-planar but p3, then the pick would result in wrong results.

                      – N.T.C
                      Jan 3 at 8:53











                    • @N.T.C Yes youre right sorry I forgot to submit the other part of answer ... silly me :)

                      – Spektre
                      Jan 3 at 8:59


















                    0
















                    1. normal n



                      pick any 3 points from your dataset that are not on a single line let call them p0,p1,p2. To improve accuracy they should be more distant to each other. Now to construct a normal vector do this:



                      n = cross( p1-p0 , p2-p0 ); // perpendicular vector to both operands of cross
                      
n /= |n|;                   // unit vector



                    2. check all points



                      for any point on the plane formed by p0,p1,p2 the altitude component (in normal direction) should be zero so for any point p:



                      |dot( p-p0 , n )|<=1e-10


                      the 1e-10 is just some zero threshold due to accuracy loss on FPU ... Any point not satisfying the condition does not belong to the plane ...




                    So how to pick 3 points forming a triangle ?





                    1. pick p0,p1



                      pick p0 as point that has minimal x,y,z coordinates and p1 that has maximal x,y,z coordinates. This ensures they are distant enough.




                    2. pick p2



                      now sear the points and find such that |dot( p1-p0, pi-p0 )| is minimal. While |pi-p0| is not zero and big enough (for example at least 0.1*|p1-p0|). That ensures the points form a triangle and are not too close to each.




                    All of this is doable in O(n) so is still fast ...






                    share|improve this answer


























                    • Isn't this same as what I described in my question? The problem here is actually what criterion to use to pick p1, p2 and p3. It is quite tricky since, for example, if all points are co-planar but p3, then the pick would result in wrong results.

                      – N.T.C
                      Jan 3 at 8:53











                    • @N.T.C Yes youre right sorry I forgot to submit the other part of answer ... silly me :)

                      – Spektre
                      Jan 3 at 8:59
















                    0












                    0








                    0









                    1. normal n



                      pick any 3 points from your dataset that are not on a single line let call them p0,p1,p2. To improve accuracy they should be more distant to each other. Now to construct a normal vector do this:



                      n = cross( p1-p0 , p2-p0 ); // perpendicular vector to both operands of cross
                      
n /= |n|;                   // unit vector



                    2. check all points



                      for any point on the plane formed by p0,p1,p2 the altitude component (in normal direction) should be zero so for any point p:



                      |dot( p-p0 , n )|<=1e-10


                      the 1e-10 is just some zero threshold due to accuracy loss on FPU ... Any point not satisfying the condition does not belong to the plane ...




                    So how to pick 3 points forming a triangle ?





                    1. pick p0,p1



                      pick p0 as point that has minimal x,y,z coordinates and p1 that has maximal x,y,z coordinates. This ensures they are distant enough.




                    2. pick p2



                      now sear the points and find such that |dot( p1-p0, pi-p0 )| is minimal. While |pi-p0| is not zero and big enough (for example at least 0.1*|p1-p0|). That ensures the points form a triangle and are not too close to each.




                    All of this is doable in O(n) so is still fast ...






                    share|improve this answer

















                    1. normal n



                      pick any 3 points from your dataset that are not on a single line let call them p0,p1,p2. To improve accuracy they should be more distant to each other. Now to construct a normal vector do this:



                      n = cross( p1-p0 , p2-p0 ); // perpendicular vector to both operands of cross
                      
n /= |n|;                   // unit vector



                    2. check all points



                      for any point on the plane formed by p0,p1,p2 the altitude component (in normal direction) should be zero so for any point p:



                      |dot( p-p0 , n )|<=1e-10


                      the 1e-10 is just some zero threshold due to accuracy loss on FPU ... Any point not satisfying the condition does not belong to the plane ...




                    So how to pick 3 points forming a triangle ?





                    1. pick p0,p1



                      pick p0 as point that has minimal x,y,z coordinates and p1 that has maximal x,y,z coordinates. This ensures they are distant enough.




                    2. pick p2



                      now sear the points and find such that |dot( p1-p0, pi-p0 )| is minimal. While |pi-p0| is not zero and big enough (for example at least 0.1*|p1-p0|). That ensures the points form a triangle and are not too close to each.




                    All of this is doable in O(n) so is still fast ...







                    share|improve this answer














                    share|improve this answer



                    share|improve this answer








                    edited Jan 3 at 8:58

























                    answered Jan 3 at 8:35









                    SpektreSpektre

                    30.3k650220




                    30.3k650220













                    • Isn't this same as what I described in my question? The problem here is actually what criterion to use to pick p1, p2 and p3. It is quite tricky since, for example, if all points are co-planar but p3, then the pick would result in wrong results.

                      – N.T.C
                      Jan 3 at 8:53











                    • @N.T.C Yes youre right sorry I forgot to submit the other part of answer ... silly me :)

                      – Spektre
                      Jan 3 at 8:59





















                    • Isn't this same as what I described in my question? The problem here is actually what criterion to use to pick p1, p2 and p3. It is quite tricky since, for example, if all points are co-planar but p3, then the pick would result in wrong results.

                      – N.T.C
                      Jan 3 at 8:53











                    • @N.T.C Yes youre right sorry I forgot to submit the other part of answer ... silly me :)

                      – Spektre
                      Jan 3 at 8:59



















                    Isn't this same as what I described in my question? The problem here is actually what criterion to use to pick p1, p2 and p3. It is quite tricky since, for example, if all points are co-planar but p3, then the pick would result in wrong results.

                    – N.T.C
                    Jan 3 at 8:53





                    Isn't this same as what I described in my question? The problem here is actually what criterion to use to pick p1, p2 and p3. It is quite tricky since, for example, if all points are co-planar but p3, then the pick would result in wrong results.

                    – N.T.C
                    Jan 3 at 8:53













                    @N.T.C Yes youre right sorry I forgot to submit the other part of answer ... silly me :)

                    – Spektre
                    Jan 3 at 8:59







                    @N.T.C Yes youre right sorry I forgot to submit the other part of answer ... silly me :)

                    – Spektre
                    Jan 3 at 8:59




















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