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Forum Maschinenbau Rostock • Thema anzeigen - 33.3

33.3

33.3

Beitragvon Marge » Fr 12. Nov 2010, 11:24

stimmt {0<= R<= 1
0<= phi<= 1
-pi/2<= t<= pi/2}
als normalbereich?
Marge
 
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Re: 33.3

Beitragvon Stephan » Fr 12. Nov 2010, 15:53

Stephan
 
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Registriert: Mo 19. Okt 2009, 11:08

Re: 33.3

Beitragvon Gunnar » Fr 12. Nov 2010, 16:21

Würde ich auch so sagen. Ich hab als Bereich 0 ≤ r ≤ 1 ʌ 0 ≤ phi ≤ 2¶ ʌ 0 ≤ teta ≤ ½¶. Ich komm damit auch auf 2¶R^3.
Gunnar
 
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Registriert: So 25. Okt 2009, 13:10
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Re: 33.3

Beitragvon @ntje » Fr 12. Nov 2010, 18:51

was tau und phi sind sieht man auch gut bei wiki unter "kugelkoordinaten" ;)
@ntje
 
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Registriert: Fr 23. Okt 2009, 11:11

Re: 33.3

Beitragvon Evolutionsbremse » Sa 13. Nov 2010, 13:38

Ich hab 2π raus.
Wenn man bei V = 2*π*r^3 - r in den Grenzen [0;1] einsetzt, erhält man 2*π.
Evolutionsbremse
 
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Registriert: Mi 4. Aug 2010, 07:24

Re: 33.3

Beitragvon Maschinenbauerin » So 14. Nov 2010, 11:45

Ich komme auch auf 2phi.
Maschinenbauerin
 
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