Re: fit plane to xz axis of data

• To: mathgroup at smc.vnet.net
• Subject: [mg89626] Re: fit plane to xz axis of data
• From: Jean-Marc Gulliet <jeanmarc.gulliet at gmail.com>
• Date: Sun, 15 Jun 2008 06:14:26 -0400 (EDT)
• Organization: The Open University, Milton Keynes, UK
• References: <g30399\$mes\$1@smc.vnet.net>

```will parr wrote:

[snip]

>> The following will find the best-fitting (i.e.,
>> least-squares) plane:
>>
>> m = Mean[data]; {u,w,v] =
>> = SingularValueDecomposition[#-m&/@data];
>>
>> v is a rotation matrix. Tr[w,List] (i.e., the
>> diagonal elements of w)
>> gives the norms of the projections of the points onto
>> the new axes.
>> They are in decreasing order, so
>>
>>   ({x,y,z}-m).v[[All,3]] == 0
>>
>> gives the equation which the best-fitting plane
>> satisfies.
>>
>
> I'm really interested in this solution, but when I to use this method, I get an answer that I don't really understand, and cannot display (in the form of a plane).
>
> using the original data in this message:
>
> In[122]:= m = Mean[data]; {u, w, v} =
>  SingularValueDecomposition[# - m & /@ data];
>
> In[123]:= Tr[w, List]
>
> Out[123]= {16.3822, 9.63356, 5.71769}
>
> In[124]:= ({x, y, z} - m).v[[All, 3]] == 0
>
> Out[124]=
> 0.984797 (-9.86988 + x) + 0.134151 (0.269052+ y) +
>   0.110358 (-1.44535 + z) == 0
>
> I don't understand this output, it is not of the form:
>
> ax + by + cz

[snip]

Solve the equation returned by ({x, y, z} - m).v[[All, 3]] == 0 for z.
For instance,

plane = z /. Solve[%, z][[1]] // Expand

returns

89.1934- 8.92363 x - 1.2156 y

Regards,
- Jean-Marc

```

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