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Forum Maschinenbau Rostock • Thema anzeigen - 33.4

33.4

33.4

Beitragvon indianahorst » Mi 10. Nov 2010, 00:12

Normalbereich:
-1 < x < 1
-1 < y < 1
0 < z < 4

M = 16

Schwerpunkt:
x_s = 0

y_s = 0

z_s = 1/16 * Integral von 0 bis 4 (20z - 4z^2)dz = 4,67 (=14/3).
Schwerpunkt ausserhalb der Pyramide?!
indianahorst
 
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Re: 33.4

Beitragvon Matze09 » Mi 10. Nov 2010, 16:14

Rechnest du mit deinem Normalbereich nicht einen Quader aus?
Man muss doch sicherlich eine Variable abgängig von einer anderen schreiben, um den "anstieg" der Pyramide mit einzubeziehen, oder täusch ich mich?

Gruß
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Re: 33.4

Beitragvon rene.zielke » Mi 10. Nov 2010, 17:44

s(0|0|0,85)
M=64/3
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Re: 33.4

Beitragvon mmlifeguard » Mi 10. Nov 2010, 18:27

Der Schwerpunkt einer geraden Pyramide teilt die gerade Senkrecht auf der Grundfläche durch die spitze im Verhältnis 3:1

demnach 1/4 * h müsste doch theoretisch 1 sein.... (0,0,1)
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Re: 33.4

Beitragvon rene.zielke » Do 11. Nov 2010, 12:12

Wir haben aber eine inhomogene Masseverteilung.
Hier speziell nimt die DIchte mit Höhe ab, wodurch sich ja logischerweise Der Schwerpunkt nach unten verschieben muss.
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Re: 33.4

Beitragvon Eule » Fr 12. Nov 2010, 17:33

Wie sieht dein Normalbereich aus?
Eule
 
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Re: 33.4

Beitragvon Rico » Fr 12. Nov 2010, 17:40

Rico
 
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Re: 33.4

Beitragvon Stephan » Fr 12. Nov 2010, 20:10

-1 < x < 1
-1 < y < 1
0 < z < ?
wo ist der ansatz bzw der denkfehler?
"Die große Mehrzahl unserer Importe kommt von außerhalb des Landes." George W. Bush
"Zwei Dinge sind unendlich, das Universum und die menschliche Dummheit, aber bei dem Universum bin ich mir noch nicht ganz sicher." Albert Einstein
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Re: 33.4

Beitragvon rene.zielke » Fr 12. Nov 2010, 22:07

Wenn d die aufgaben mit normalbereichen lösen wisst musst du:
-1+0,25zx1-0,25z
-1+0,25zy<1-0,25z
0z<
nehmen, weil die Pyramide nach oben hin zuläuft; deines entspräche einem Quader.
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Re: 33.4

Beitragvon indianahorst » So 14. Nov 2010, 14:21

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