Pista, Bo, Ric, Chris, Brian --
Thanks again for your responses! I got the interfacial angles
(after expressing every planar angle as internal to its face).
With the planars grouped into per-vertex triples, edge names
defined for output labelling, and the triples each fed to
IFA =: 3 : 0
for_I. i.3 do.
'a b c' =. I |. y.
RAD =. _2 o. ((2 o.a) - (2 o.b)*2 o.c) % (1 o.b)*1 o.c
((>I{2|.EDGES),' : ',(":RAD, dfr RAD),CRLF) fappend OUT
end.
, the 21 interfacials came out sensibly -- once rounded to
5th place, 12 unique radians (for 9 duals and 3 singles).
Best regards,
Pete
PMA wrote:
> Hi Pista!
>
> This comes just in time to vet my results, got for better
> or worse after a Bo > Spherical Trig > Wolfram search.
>
> At first glance I see a slight difference in our formulae,
> but won't ask about it yet (esp. lest it prove cosmetic).
> Comparing further....
>
> You've gone to a lot of trouble here. Thank you!
>
> Pete
>
>
> Istvan Kadar wrote:
>> Hi Pete,
>>
>> A NUMERICAL EXAMPLE
>>
>> NB. A B
>> known NB. \ /
>> edge-angles NB. \ /
>> a=:57 01 45.2 NB. \ c /
>> b=:87 42 41.1 NB. \ /
>> c=:55 37 19.6 NB. \ /
>> NB. D
>> unknown NB. a | b
>> face-angles NB. |
>> C=: ? NB. |
>> A=: ? NB. |
>> B=: ? NB. |
>> NB. C
>>
>> a + b + c << 360
>> (57 01 45.2+87 42 41.1+55 37 19.6=200 21 45.9<<360)
>>
>> A formula for the solution
>>
>> cos_C =: ((cos c)-~(cos a)*cos b)%(-sin a)*sin b
>> C =: acos cos_C
>>
>> ]sin_a=:1&o.rfd((%60)p.~a)
>> 0.838948
>> 9j7":sin_a
>> 0.8389482
>> ]cos_a=:2&o.rfd((%60)p.~a)
>> 0.544211
>> 9j7":cos_a
>> 0.5442112
>>
>> ]sin_b=:1&o.rfd((%60)p.~b)
>> 0.999202
>> 9j7":sin_b
>> 0.9992024
>> ]cos_b=:2&o.rfd((%60)p.~b)
>> 0.0399327
>> 9j7":cos_b
>> 0.0399327
>>
>> ]cos_c=:2&o.rfd((%60)p.~c)
>> 0.564649
>> 9j7":cos_c
>> 0.5646485
>>
>> ]cos_C =: ((cos_c)-~cos_a*cos_b)%-sin_a*sin_b
>> 0.647656
>> 9j7":cos_C =: ((cos_c)-~cos_a*cos_b)%-sin_a*sin_b
>> 0.6476563
>> _2&o.0.6476563
>> 0.866292
>> dfr=:rfd^:_1
>> 9j7":_2&o.0.6476563
>> 0.8662919 radian
>> dfr 0.8662919
>> 49.6349
>>
>> C=:49.6349 degree (49° 38' 05.5" )
>>
>> Best regards
>> Pista
>>
>> 2009/8/22 PMA <[email protected]>
>>
>>> Bo, Thank you for these leads!
>>>
>>> I felt the algebra would be just plain,
>>> but hadn't the grasp to specify it.
>>> Will pursue....
>>>
>>> Best regards,
>>> Pete
>>>
>>>
>>> Bo Jacoby wrote:
>>>> Isn't it just plain linear algebra? You have a 3-corner and knows
>>>> the 3 cosines of angles between 3 vectors. Compute the cosines
>>>> of angles between pairwise normals to the vectors.
>>>>
>>>> Venlig hilsen, Bo.
>>>>
>>>>
>>>> --- Den fre 21/8/09 skrev Sherlock, Ric <[email protected]>:
>>>>
>>>>> Fra: Sherlock, Ric <[email protected]>
>>>>> Emne: Re: [Jprogramming] Szilassi's toroidal heptahedron: interfacial
>>> angles? Last try.
>>>>> Til: "Programming forum" <[email protected]>
>>>>> Dato: fredag 21. august 2009 03.07
>>>>> Hi Peter,
>>>>> You could try providing some sample data and show what you
>>>>> expect the result to be.
>>>>> That might elicit a better response.
>>>>>
>>>>>> From: PMA
>>>>>>
>>>>>> Here is a final crack at posing my question -- this is
>>>>> as whittled as
>>>>>> I can whittle:
>>>>>>
>>>>>> Given a 3-D vertex of 3 faces with all edge-to-edge
>>>>> angles known,
>>>>>> how does one calculate its face-to-face angles (all,
>>>>> one per edge)?
>>>>>> If there come to mind other sites especially
>>>>> interested in this sort
>>>>>> of query, I'd appreciate hearing.
>>>>>>
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>>>>>
>>>>
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>>> tilbud på LCD TV! Se her http://dk.yahoo.com/r/pat/lcd
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>>>>
>>>
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>>
>
>
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