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Expert
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Apr 20, 2011, 05:42 AM
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Originally Posted by abver
surface area = pi * r 2 + r * pi * h for this figura
Sounds once again like you're talking about an open-ended cylinder, whose surface area consists of the cylinder wall plus one end cap:
And the volume would be (as Unknown008 noted, oh so long ago):
So, abver, did you ever figure out where the factor 3.3629 came from in your original post?
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Junior Member
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Apr 20, 2011, 09:40 PM
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Comment on ebaines's post
serb : V = pi * r ^ 2 * h / 3,3629.. A = pi * r ^ 2 + pi * r * h su obrasci koji definisu jednu geometrisku figuru ( koja nije valjak) to jest njenu zapreminu I povrsinu a pitanje je koja je to figura, I iz cega se sastoji ? (3,3629.. jnaravno da sam shvatio odakle potice, mogao bi da objasnim, al mi se ne gubi vreme) ustvari mene zanima da li su ovi obrasci poznati vrsnim ekspertima kao sto ste vi I uopste (inace meni je matematika hobi ali samo ponekad)
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Expert
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Apr 21, 2011, 05:45 AM
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abver's response as translated by Google:
"the patterns that define a geometric figure (not roller) that is its volume and surface area and the question is what that figure, and from what constitutes? (3.3629 .. jnaravno I figured originated, could explain, but I do not waste time) I really care whether these patterns are known excellent experts like you, and in general (who is my math is a hobby, but only sometimes)."
Wow, that's really helpful.
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Uber Member
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Apr 21, 2011, 09:36 AM
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Indeed, this sort of hobby, I dare say is not healthy at all! D:
If at least you are deriving that formula, with meaningful equations and such, then it's okay, but here... here... I just don't see the point in that, sorry. And I don't think that anyone who has math as a hobby would like to know those patterns.
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Junior Member
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Apr 21, 2011, 02:04 PM
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Comment on Unknown008's post
Serb : ako mozes da razumes zasto je (opposite AB / sec) + ((opposite AB - (opposite AB / sec)) * 4/3) ~ arc AB mozda I mo nadjes poentu u gornjim jednacinama
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Junior Member
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Apr 21, 2011, 02:18 PM
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Comment on Unknown008's post
pod uslovom da je hipotenuza = precnik
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Ultra Member
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Apr 21, 2011, 02:36 PM
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Jjuiibn;j h hmnk mxvnncvuygaefnbamn
I checked Google and this "problem" has been posted on a dozen different sites. No one else appears to know what it means either.
Especially, the 3363.
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Junior Member
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Apr 21, 2011, 06:41 PM
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Comment on galactus's post
2 pi 2 / (pi 2 - 4) je formula koja daje odnos zbira povrsina svih jedinstvenih segments (pi 2 - 4) I zbira odgovarajuceg broja povrsina circle (2 pi 2), odnosno volume of a cylinder I volume geometriskog tela koja se sastoji od povrsina svih jedinstvenih segments a taj odnos je 3,3629.. a odgovor je (diameter * 1/2 chord A) / 2 = triangle A ;; dakle (r 2 * pi / 2 ) - triangle = area of a two segments ;; dakle triangle A / chord A = x ;; dakle (r 2 * pi / 2) / x = diameter * 1/2 pi ;; dakle ((diameter * 1/2 pi) - chord A) * x = (r 2 * pi / 2) - triangle A = area of a two segments;; dakle zbir svih jedinstvenih chords je RADIUS a zbir odgovarajuceg broja diameters je (2 r * pi) / 4 ;; dakle ((2 r * pi) / 4 * 1/2 pi - r) * x = zbir povrsina svih jedinstvenih segments a zbir odgovarajuceg broja povrsina circle je ((2 r * pi) / 4 * 1/2 pi * 2) * x ;; dakle (2 r * pi / 4 * pi) / (2 r * pi / 4 * 1/2 pi - r) ova formula moze da se svede na 2 pi 2 / (pi 2 - 4)
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Uber Member
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Apr 22, 2011, 08:55 AM
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Okay, I can't understand what you're talking about. Give me a practical example where this is used.
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Junior Member
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Apr 24, 2011, 02:41 PM
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Comment on Unknown008's post
OK , pokusacu da objasnim , nacrtaj cetvrtinu kruga, sada zamisli da je centralna tacka povezana poluprecnicima (radius you mnozini) sa svakom tackom na cetvrtini kruznice, na taj nacin dobio bi poluprecnike you svim uglovima od 0 do 90, sada zamisli da su svi ti poluprecnici povezani you nizu redom od vertikalnog do horizontalnog, tako da zadnja tacka jednog poluprecnika predstavlja pocetnu tacku drugog (sledeceg) poluprecnika, na taj nacin dobio bi poligon sa beskonacno stranica, sada zamisli da svaku stranicu (poluprecnik) zamenjuje tacka, na taj nacin dobio bi iskrivljenu liniju koja bi bila identicna cetvrtini kruznice. Vracamo se na pocetak, zamisli da svaki poluprecnik predstavlja hipotenuzu, you tom slucaju zbir svih stranica (opposite I adjancent) pravouglih trouglova predstavljala bi pravu does, koja bi bila jednaka poluprecniku (radius) kruga cija je cetvrtina nacrtana na pocetku, ako ovo razumes ti napisi
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Junior Member
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Apr 28, 2011, 03:29 PM
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Comment on galactus's post
Sustina je da, kada od povrsine polukruga oduzmes pravougli trougao ciji je poluprecnik jednak hipotenuzi, dobices povrsinu dva segments, e sada kada bi oduzeo sve moguce pravougle trouglove od odgovarajuceg broja povrsina polukruga, dobio bi zbir svih segments, sto objasnjavaju gornje formule (nije valjda tolko komplikovano)
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Junior Member
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Jun 20, 2011, 11:48 AM
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Comment on Unknown008's post
ideja oko normalne proekcije tacke sa kruznice na poluprecnik, gde ja odnos te, tacke / njene proekcije na radius = precnik / tetive (mat pojam) paralelne sa tom tackom. Ima za posledicu mnogo efekata koji ako se pravilno razumu mogu da daju dosta rezultata. Medjutim najvazniji predstavja niz kojim se izracunava funkcija sinus, gde svaki deo niza similuje (glumi) tu proekciju. Zatim da projektovana tacka ustvari konvertuje pravougli (jednakostranicni) trougao ali I u isto vreme ona konvertuje I prostor, I to ne jedan vec beskonacno mnogo razlicitih prostora od kojih se svaki sastoji od povrsina pravouglih (jednakostranicnih) trouglova. Zatim da se povrsina polukruga moze razumeti kao zbir svih kvadrata razlicitih tetiva. Jedna od posledica je I gornji obrazac Medjutim sve ovo funkcijonise samo kod KRUGA I nikakvoj drugoj varijanti I samim tim ova ideja NE MOZE da se iskoristi za neku opstu teoriju. Ovih dana vidjam na ozbiljnim sajtovima, pogresne interpretacije ( zabrinjavajuce )
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Junior Member
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Jun 20, 2011, 01:01 PM
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Comment on Unknown008's post
Napravio sam gresku u gornjem komentaru mislio sam na jednakokraki trougao, ne na jednakostranicni trougao
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Senior Member
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Jun 20, 2011, 07:12 PM
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At least now I know how to say "isosceles triangle" in Serbian. :)
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