sábado, 11 de enero de 2020

números fraccionaris binaris

(0.01)_{b} = (01/100)_{b} = (1/4)
(0.10)_{b} = (10/100)_{b} = (2/4) = (1/2)
(0.11)_{b} = (11/100)_{b} = (3/4)


(0.001)_{b} = (001/1000)_{b} = (1/8)
(0.010)_{b} = (010/1000)_{b} = (2/8) = (1/4)
(0.011)_{b} = (011/1000)_{b} = (3/8)
(0.100)_{b} = (100/1000)_{b} = (4/8) = (1/2)
(0.101)_{b} = (101/1000)_{b} = (5/8)
(0.110)_{b} = (011/1000)_{b} = (6/8) = (3/4)
(0.111)_{b} = (111/1000)_{b} = (7/8)


(100)_{b}·x+(-1)·x=(11)_{b}·x
(0.01...)_{b} = (1/3)
(0.10...)_{b} = (2/3)
(0.11...)_{b} = (3/3)


(1000)_{b}·x+(-1)·x=(111)_{b}·x
(0.001...)_{b} = (1/7)
(0.010...)_{b} = (2/7)
(0.011...)_{b} = (3/7)
(0.100...)_{b} = (4/7)
(0.101...)_{b} = (5/7)
(0.110...)_{b} = (6/7)
(0.111...)_{b} = (7/7)


(0.0011...)_{b} = (3/15) = (1/5)
(0.0110...)_{b} = (6/15) = (2/5)
(0.1001...)_{b} = (9/15) = (3/5)
(0.1100...)_{b} = (12/15) = (4/5)
(0.1111...)_{b} = (15/15) = (5/5)


(0.000111...)_{b} = (7/63) = (1/9)
(0.001110...)_{b} = (14/63) = (2/9)
(0.010101...)_{b} = (21/63) = (3/9) = (1/3)
(0.011100...)_{b} = (28/63) = (4/9)
(0.100011...)_{b} = (35/63) = (5/9)
(0.101010...)_{b} = (42/63) = (6/9) = (2/3)
(0.110001...)_{b} = (49/63) = (7/9)
(0.111000...)_{b} = (56/63) = (8/9)
(0.111111...)_{b} = (63/63) = (9/9)


(0.1)_{b}·(0.0011...)_{b} = (1/10)
(0.1)_{b}·(0.0110...)_{b} = (2/10)
(0.1)_{b}·(0.1001...)_{b} = (3/10)
(0.1)_{b}·(0.1100...)_{b} = (4/10)
(0.1)_{b}·(0.1111...)_{b} = (5/10) = (1/2)
(0.1)_{b}·(1.0011...)_{b} = (6/10)
(0.1)_{b}·(1.0110...)_{b} = (7/10)
(0.1)_{b}·(1.1001...)_{b} = (8/10)
(0.1)_{b}·(1.1100...)_{b} = (9/10)
(0.1)_{b}·(1.1111...)_{b} = (10/10)


(1/10)+1 = (0.1)_{b}·(10.0011...)_{b}
(1/10)+2 = (0.1)_{b}·(100.0011...)_{b}
(1/10)+3 = (0.1)_{b}·(110.0011...)_{b}
(1/10)+4 = (0.1)_{b}·(1000.0011...)_{b}


1+(-1)·(1/10) = (0.1)_{b}·(01.1100...)_{b}
2+(-1)·(1/10) = (0.1)_{b}·(011.1100...)_{b}
3+(-1)·(1/10) = (0.1)_{b}·(101.1100...)_{b}
4+(-1)·(1/10) = (0.1)_{b}·(0111.1100...)_{b}


(0.1)_{b}·(0.01...)_{b} = (1/6)
(0.1)_{b}·(0.10...)_{b} = (2/6) = (1/3)
(0.1)_{b}·(0.11...)_{b} = (3/6) = (1/2)
(0.1)_{b}·(1.01...)_{b} = (4/6) = (2/3)
(0.1)_{b}·(1.10...)_{b} = (5/6)
(0.1)_{b}·(1.11...)_{b} = (6/6)


(0.0001011101...)_{b} = (93/1023) = (1/11)

assembler x=x

x=x
{
mov bx,x
mov ax,[bx]
mov [bx],ax
}


x=not(x)
{
mov bx,x
mov ax,[bx]
not ax
mov [bx],ax
}


x=not(not(x))
{
mov bx,x
mov ax,[bx]
not ax
not ax
mov [bx],ax
}

grafic m minuscula estrident

M-Minuscula( int n-x , int n-y , int x , int y )
{


for( k=1 ; k [< (n-x)  ; k++ )
{
m-k = k;
m-x = n-x;
x = x;
y = y;
put-pixel-color-x( color , x+( (n-x)+not(m-k) ) , y+( (m-x)+not(k) ) )
}
for( k=(-1) ; k >] not(n-x)  ; k-- )
{
m-k = not(k);
m-x = not(n-x);
x = not(x);
y = not(y);
put-pixel-color-y( color , not(x)+( not(n-x)+(m-k) ) , not(y)+( not(m-x)+k ) )
}


for( k=1 ; k [< (n-y)  ; k++ )
{
m-k = k;
m-y = n-y;
x = x;
y = y;
put-pixel-color-x( color , x+( (n-x)+(m-k) ) , y+( (m-y)+not(k) ) )
}
for( k=(-1) ; k >] not(n-y)  ; k-- )
{
m-k = not(k);
m-y = not(n-y);
x = not(x);
y = not(y);
put-pixel-color-y( color , not(x)+( not(n-x)+not(m-k) ) , not(y)+( not(m-y)+k ) )
}


}

funció botó circular y coordenades polars inicialització

for( radi = 1 ; radi [< m ; radi++ )
{
for( angle = 1 ; angle [< 360 ; angle++ )
{
radi-x = radi;
radi-y = radi;
z = y;
angle = angle;
botó-x-positiu[angle][radi-x] = x + radi·cos(angle);
botó-y-positiu[angle][radi-y] = z + radi·sin(angle);
botó-de-circunferencia-positiu[angle][radi] = angle·radi;
botó-de-sector-circular-positiu[angle][radi] = (1/2)·angle·(radi·radi);
}
}


for( radi = not(1) ; radi >] not(m) ; radi-- )
{
for( angle = not(1) ; angle >] not(360) ; angle-- )
{
radi-x = not(radi);
radi-y = not(radi);
z = not(y);
angle = not(angle);
botó-x-negatiu[not(angle)][radi-x] = not(x) + not(radi)·cos(not(angle));
botó-y-negatiu[not(angle)][radi-y] = not(z) + not(radi)·sin(not(angle));
botó-de-circunferencia-negatiu[not(angle)][not(radi)] = not(angle)·not(radi);
botó-de-sector-circular-negatiu[not(angle)][not(radi)] = (1/2)·not(angle)·(not(radi)·not(radi));
}
}

viernes, 10 de enero de 2020

mecànica de colisió de una partícula amb un extrem de una barra

txoc de una partícula en un extrem de una barra a velocitat constant V:
m_{1} = k·m_{2}


m_{1}·d_{t}[x(t_{0})] = m_{1}·d_{t}[x(t_{1})] + m_{2}·d_{t}[s(t_{2})]·R


m_{1}·V = m_{1}·d_{t}[x(t_{1})] + m_{2}·d_{t}[s(t_{2})]·R


txoc inelástic:
si d_{t}[x(t_{1})] = d_{t}[s(t_{2})]·R ==>


d_{t}[x(t_{1})] = V·( m_{1}/(m_{1}+m_{2}) )


d_{t}[x(t_{1})] = V·( k/(k+1) )
d_{t}[s(t_{2})] = (V/R)·(k/(k+1) )


txoc elástic


m_{1}·V = m_{1}·(-V) + m_{2}·d_{t}[s(t_{2})]·R


d_{t}[s(t_{2})] =  (V/R)·(m_{1}/m_{2})
d_{t}[s(t_{2})] =  (V/R)·k

suma directa y números

3 = x^{2}+y^{2}
3 = (1/2)·( x[+]y )^{2}
6 = ( x[+]y )^{2}
6^{(1/2)} = x[+]y


x = ( 6^{(1/2)}/2 )
y = ( 6^{(1/2)}/2 )


5 = x^{2}+y^{2}
5 = (1/2)·( x[+]y )^{2}
10 = ( x[+]y )^{2}
10^{(1/2)} = x[+]y


x = ( 10^{(1/2)}/2 )
y = ( 10^{(1/2)}/2 )


2 = x^{2}+y^{2}
2 = (1/2)·( x[+]y )^{2}
4 = ( x[+]y )^{2}
4^{(1/2)} = x[+]y
2 = x[+]y


x = 1
y = 1

suma directa y ecuacions cuártiques


x^{4}+y^{4} = k
(1/8)·( x[+]y )^{4} = k
( x[+]y )^{4} = 8k
( x[+]y ) = 8^{(1/4)}·k^{(1/4)}


x = (4/8^{(3/4)})·k^{(1/4)}
y = (4/8^{(3/4)})·k^{(1/4)}


x^{4}+y^{4} = (x+y)
(1/8)·( x[+]y )^{4} = (x+y)
( x[+]y )^{4} = 8·(x+y)
( x[+]y ) = 8^{(1/4)}·(x+y)^{(1/4)}


x = 1
y = 1


x^{4}+y^{4} = (x+y)^{3}
(1/8)·( x[+]y )^{4} = (x+y)^{3}
( x[+]y )^{4} = 8·(x+y)^{3}
( x[+]y ) = 8^{(1/4)}·(x+y)^{(3/4)}


x = 4
y = 4