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Socratis
May 24, 2012, 10:19:03 AM5/24/12
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--- o=0,01=cm --Intergrale del cono r=6o,h=8o.
o^3=o=(r^2*h*pi)/(3h^2)
o^3=o=((6o)^2*8o*pi)/(3*(8o)^2) = 4.71238898o^3
.
\/\/\/\/\/\/\/\/ 15o *o = 70.68583471 cm^3
\/\/\/\/\/\/\/ 13o *o = 61.26105674 cm^3
\/\/\/\/\/\/ 11o *o = 51.83627878 cm^3
\/\/\/\/\/ 9o *o = 42.41150082 cm^3
\/\/\/\/ 7o *o = 32.98672286 cm^3
\/\/\/ 5o *o = 23.56194490 cm^3
\/\/ 3o *o = 14.13716694 cm^3
\/ o = 4.712388980 cm^3
V.cono. (o,8o)= o* h^2 = 301.5928947cm^3. = 3.01592 dl.
S.int. (i , 8i) = i * h^2 = 301.5928947dm^3 = 301.592 Litre
S.Int. (1 , 8) = 1*h^2 = 301.5928947m^3 = 301592,8 Litre
Che in pratica vuol dire che la punta del cono ha un volume
pari a 4.71.cm^3, come per i vale 4.71..dm^3
come per 1 vale 4.71..m^3.
Come per o =mm vale 4.71..mm^3,
e che in totale esistono h^2 conetti, distribuiti su 8 fette punta
compresa.
Socratis.
Socratis
May 27, 2012, 11:03:10 AM5/27/12
to
--- o=0,01=cm --Intergrale du conus, r=6o,h=8o.
Formul pour trouver V =o^3=o --> (r^2*h*pi)/(3h^2)
V=o---> ((6o)^2*8o*pi)/(3*(8o)^2) = 4.71238898o^3
Formul pour trouver V =o^3=o --> (r^2*h*pi)/(3h^2)
V=o---> ((6o)^2*8o*pi)/(3*(8o)^2) = 4.71238898o^3
.
\/\/\/\/\/\/\/\/ 15o *o = 70.68583471 cm^3
\/\/\/\/\/\/\/ 13o *o = 61.26105674 cm^3
\/\/\/\/\/\/ 11o *o = 51.83627878 cm^3
\/\/\/\/\/ 9o *o = 42.41150082 cm^3
\/\/\/\/ 7o *o = 32.98672286 cm^3
\/\/\/ 5o *o = 23.56194490 cm^3
\/\/ 3o *o = 14.13716694 cm^3
\/ o = 4.712388980 cm^3
V.cono. (o,8o)= o* h^2 = 301.5928947cm^3. = 3.01592 dl.
S.int. (i , 8i) = i * h^2 = 301.5928947dm^3 = 301.592 Litre
S.Int. (1 , 8) = 1*h^2 = 301.5928947m^3 = 301592,8 Litre
Socratis.
\/\/\/\/\/\/\/\/ 15o *o = 70.68583471 cm^3
\/\/\/\/\/\/\/ 13o *o = 61.26105674 cm^3
\/\/\/\/\/\/ 11o *o = 51.83627878 cm^3
\/\/\/\/\/ 9o *o = 42.41150082 cm^3
\/\/\/\/ 7o *o = 32.98672286 cm^3
\/\/\/ 5o *o = 23.56194490 cm^3
\/\/ 3o *o = 14.13716694 cm^3
\/ o = 4.712388980 cm^3
V.cono. (o,8o)= o* h^2 = 301.5928947cm^3. = 3.01592 dl.
S.int. (i , 8i) = i * h^2 = 301.5928947dm^3 = 301.592 Litre
S.Int. (1 , 8) = 1*h^2 = 301.5928947m^3 = 301592,8 Litre
Socratis
May 27, 2012, 2:25:00 PM5/27/12
to
--- dy=o=0,01=cm --Intergral de conus : r=6.o.x, h=8.o.y.
Formula per trovare V = o^3=dy= o --> r^2*pi)/3h
V=o^3---> ((6.o.x)^2*pi)/3*8.o.y = 4.71238898o^3
.
\/\/\/\/\/\/\/\/ 15o *o = 70.68583471 cm^3, Sum.int=301.59..
si trouve par : (r^2*pi)/3*h)*(n.dh)^2
ex : int \0.5o\ = (4.71..)*(5o)^2 =117. 8097245.. o^3
Socratis.
Formula per trovare V = o^3=dy= o --> r^2*pi)/3h
V=o^3---> ((6.o.x)^2*pi)/3*8.o.y = 4.71238898o^3
.
\/\/\/\/\/\/\/\/ 15o *o = 70.68583471 cm^3, Sum.int=301.59..
\/\/\/\/\/\/\/ 13o *o = 61.26105674 cm^3
\/\/\/\/\/\/ 11o *o = 51.83627878 cm^3
\/\/\/\/\/ 9o *o = 42.41150082 cm^3 Sum.int=117. 8097..
\/\/\/\/\/\/ 11o *o = 51.83627878 cm^3
\/\/\/\/ 7o *o = 32.98672286 cm^3
\/\/\/ 5o *o = 23.56194490 cm^3
\/\/ 3o *o = 14.13716694 cm^3
\/ o = 4.712388980 cm^3
La Sum. integral du conus, de 0 a dy=dh
\/\/\/ 5o *o = 23.56194490 cm^3
\/\/ 3o *o = 14.13716694 cm^3
\/ o = 4.712388980 cm^3
si trouve par : (r^2*pi)/3*h)*(n.dh)^2
ex : int \0.5o\ = (4.71..)*(5o)^2 =117. 8097245.. o^3
Socratis.
Socratis
May 27, 2012, 4:05:03 PM5/27/12
to
Ora e' semplicissimo, spero.......
--- dy=1dz=cm --Intergrale del cono r=6x, h=8dz.
Formula per trovare V.T,cono. = (r^2*pi*8dz)/3
V=1dv---> (6x*6y*pi)/3*8dz = 4.71238898 cm^3
.
\/\/\/\/\/\/\/\/ 15V *dz = 70.68583471 cm^3, V.T.=301.59..cm^3
\/\/\/\/\/\/\/ 13V*dz = 61.26105674 cm^3
\/\/\/\/\/\/ 11V*dz = 51.83627878 cm^3
\/\/\/\/\/ 9V*dz = 42.41150082 cm^3 Sum.int.\0, 5dz\ = @
\/\/\/\/ 7V*dz = 32.98672286 cm^3
\/\/\/ 5V*dz = 23.56194490 cm^3
\/\/ 3V*dz = 14.13716694 cm^3
\/ V*dz = 4.712388980 cm^3
Naturalmente, la somma integrale del cono, da 0 a dy=dh=dz
si trova per (r^2*pi)/3*z)*(n)^2*dz
es : int \0, 5dz\ = (V)*5^2dz =117. 8097245.. cm^3 @
Spero di non aver commesso, errore, omissione)....
nel dubbio correggetemi. Grazie.Socratis.
YBM
May 27, 2012, 8:37:13 PM5/27/12
to
Le 27.05.2012 22:05, Socratis a écrit :
> Ora e' semplicissimo, spero.......
>
> --- dy=1dz=cm --Intergrale del cono r=6x, h=8dz.
Mais nous savons que tu es intégralement con, Sorrentino/Socratis,
> Ora e' semplicissimo, spero.......
>
> --- dy=1dz=cm --Intergrale del cono r=6x, h=8dz.
il n'est nul besoin d'insister, sinistre crétin.
Socratis
May 29, 2012, 1:02:37 PM5/29/12
to
"YBM" <ybm...@nooos.fr.invalid> ha scritto nel messaggio
> Le 27.05.2012 22:05, Socratis a écrit :
>> Ora e' semplicissimo, spero.......
>>
>> --- dy=1dz=cm --Intergrale del cono r=6x, h=8dz.
>
Ce que suive, n'est pas pur les savantes.
La Tunze est plus facile a faire, que a l'explicher... :
In realte' les triangules sont une rappresentation des volumes.
dv in fonction de h^2
ex.1; r.c = 2cm e h.c = 3cm--> r.c=20mm , h.c =30mm.
V.t.c. = 400pi/3*30mm=12566.37061..mm^3
(12566.37061..mm^3)/9cm = (1396.263402mm^3)/cm = dv/cm
1cm^3 =1000mm^3 retrasformer 1000mm^3 in 1cm^3.
VVV 5*dv = 6.981317....cm^3
VV 3*dv= 4.188790205..cm^3
V dv = 1.396263402..cm^3
::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::::
ex.2 : r.c=2cm, h.c = 4cm
(V.t.c / h^2) = dv
V.t.c. = (20mm)^2*pi)/3*40mm =15755.16..mm^3
(15755.16..mm^3 /16) = dv = 1047.19..mm^3 /cm.
dv/cm, retrasforme' in cm^3/cm : dv =1.047..cm^3/cm.
VVVV 9dv
VVV 5dv
VV 3dv
V dv = 1.047...cm^3 .
V.t.c. = dv*h^2 = 1.047..cm^3*16=15.75516cm^3
Ou 16 sont les dv= triangoles que rappresentent l'integrale
cubic, de 0 a 4.
----- (integrale de 0 a 3cm . r.c=2cm, h.c = 4cm ) =
= dv*9-->1.047*9 = 9.4247..cm^3.
Bonjour. Tunze.
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