User talk:Maher ezeden aldaher

superellipse;superellipsoid;superquadrics
Area&Perimeter for Superellipse,Hypoellipse a≥b,  ∞≥ n ≥  1  ,    x^n/a^n +y^n/b^n =1 L=a+b((b((2.5)/(n+0.5))^(1/n)+(0.566(n-1)a)/n^2 )/(b+(4.5a)/(n^2+0.5))) Lx=x+(b-y)(((b-y) ((2.5)/(n+0.5))^(1/n)+(0.566(n-1)x)/n^2 )/(b-y+((4.5a)/(n^2+0.5))x)) ((b-y)/b)^(((n-1)/3) ) P=L*4 L(a-x)=L-Lx A=a b(〖0.5)〗^(n^(-1.52) ) At=A*4 A(a-x) = ay((〖0.5)〗^(n^(-1.52) )- n^4/(1+n^4 ) (x/a)+(n-1)(〖1/(n+1))〗^(n-1) (〖x/a)〗^(2n-1)- (n-1)0〖.117〗^(n-1) 〖 (x/a)〗^(2n-1)) Ax = A – A(a-x) Other forms A=a b(1-〖1/〖2n〗^1.22 )〗^ By using Binomial nested series for expanding f(x) Ax =b(x-〖 a/(n(n+1))〗^ (〖x/a)〗^(n+1) + 〖 a(1-n)/(〖2n〗^2 (2n+1) )〗^ (〖x/a)〗^(2n+1)- 〖 a(1-n)(1-2n)/(〖6n〗^3 (3n+1) )〗^ (〖x/a)〗^(3n+1)+ 〖 a(1-n)(1-2n)(1-3n)/(〖24n〗^4 (4n+1) )〗^ (〖x/a)〗^(4n+1)- …………) Ax =bx+ab∑_(m=1)^(m=∞)▒(-1)^m 〖 (∏_(m=1)^m▒(1-(m-1)n) )/(m〖!n〗^m (mn+1) )〗^ (〖x/a)〗^(mn+1) A=ab(1- 1/(n(n+1))+ ((1-n))/(〖2n〗^2 (2n+1) )-  (1-n)(1-2n)/(〖6n〗^3 (3n+1) )+ (1-n)(1-2n)(1-3n)/(〖24n〗^4 (4n+1) ) - ……) Area&Perimeter for General Asteroid 0b       General Asteroid N= 2/3     , a=b        special case of Asteroid - four cusped hypocycloid , tetracuspid, cubocycloid,       paracycle (x/a)^N+(y/b)^N=1 To find perimeter and area of general Asteroid ,first convert  the power N  to another similar superellipse curve of power n. For 0< N <1, values of n estimated as:

1.00	0.84	0.7	0.667	0.6	0.562	0.5	0.375	0.33	0.29	0.198	N 1.00	1.2	1.49	1.58	1.82	2.0	2.4	4.0	5.3	7.0	20.0	n by using Find graph program trial version  2.411 the executed relations eq.(2-8) and eq.(3-8)  as follows: n=((1.88262-4.12757N+5.3145N^2-2.21075N^3 ) )/(N-0.136122     ) N=((0.501489+0.257547n-0.0084194n^2+0.000203224n^3 ) )/(n-0.24921                                                                                                                                                       x  coordinate of astroid= a - x  coordinate of superellipse y  coordinate of astroid= b - y  coordinate of superellipse L  asteroid = L  superellipse                                                                                          L=a+b((b((2.5)/(n+.5))^(1/n)+(0.566(n-1)a)/n^2 )/(b+(4.5a)/(n^2+0.5)))                                                                              Lx   asteroid = L(a-x) superellipse                                                                                      a-x of asteroid = x1                                                                                                         L(x1)=x1+(b-y1)(((b-y1) ((2.5)/(n+.5))^(1/n)+(0.566(n-1)x1)/n^2 )/(b-y1+((4.5a)/(n^2+0.5))x1)) ((b-y1)/b)^(((n-1)/3) ) Lx  asteroid = L - Lx1 P=L*4 A asteroid = a b – a b(〖0.5)〗^(n^(-1.52) )                                                                            (11-8) At asteroid = A asteroid *4 A(x1) = ay1((〖0.5)〗^(n^(-1.52) )- n^4/(1+n^4 ) (x1/a)+(n-1)(〖1/(n+1))〗^(n-1) (〖x1/a)〗^(2n-1)-  (n-1)0〖.117〗^(n-1) (〖x1/a)〗^(2n-1)) Ax asteroid= bx - A(x1) Surface area & Volume for Superquadrics This is a general approximation to find surface area for the general equation (x/a)^n+(y/b)^n+(z/c)^n=1 that I have found based on Herons formula for n=1,Knud Thomson for n=2 ,geometry for n= infinity. Total Area=8*Q*[(a*b)^K+(a*c)^K+(b*c)^K]^(1/K),where Q=(0.5)^(n^-1.58) K=(2)^(n^-.5462) for a=b=c=r Total Area=8*Q*[(r)^2K+(r)^2K+(r)^2K]^(1/K)

This an eq.to approximate volume error%(0-2.2)for the the general superquadric with power n volume= 8*a*b*c*(1/6)^(n^-1.479) for a=b=c=r volume = 8*r^3*(1/6)^(n^-1.479) Total surface area & volume for astroidoid (x/a)^1/N+(y/b)^ 1/N +(z/c)^ 1/N =1 For 0< N <1, values of n estimated as:

1.00	0.84	0.7	0.667	0.6	0.562	0.5	0.375	0.33	0.29	0.198	N 1.00	1.2	1.49	1.58	1.82	2.0	2.4	4.0	5.3	7.0	20.0		n by using Find graph program trial version  2.411 the executed relations as follows:

n=((1.88262-4.12757N+5.3145N^2-2.21075N^3 ) )/(N-0.136122     ) N=((0.501489+0.257547n-0.0084194n^2+0.000203224n^3 ) )/(n-0.24921     ) Total Area=8*Q*[(a*b)^K+(a*c)^K+(b*c)^K]^(1/K),where Q=(0.5)^(n^-1.58) K=(2)^(n^-.5462) volume= 8*a*b*c*[1-(1/6)^(n^-1.479)]

By using pappus theorem you can find surface area and volume for superellipsoid of revolution also astroid of revolution (x/a)^n+(y/b)^n+(z/b)^n=1;r=b S=arc length of half superellipse=2*(a+b*(((2.5/(n+0.5))^(1/n))b+a(n-1)0.566/n^2)/(b+a(4.5/(0.5+n^2))) ) A=area of half superellipse = 2*(a*b*((0.5)^((n^(-1.52)))) ) surface area= 2*pi*S*r volume=2*pi*A*r Also you can apply for astroid of revolution to find surface area & volume after finding related power N to n (x/a)^1/N+(y/b)^ 1/N +(z/b)^ 1/N =1;r=b S=arc length of half superellipse=2*(a+b*(((2.5/(n+0.5))^(1/n))b+a(n-1)0.566/n^2)/(b+a(4.5/(0.5+n^2))) ) surface area = 2*pi*S*r A=area of half astroid = 2*(a*b*(1-(0.5)^((n^(-1.52)))) ) volume=2*pi*A*r