Practica 1
Páginas: 3 (527 palabras)
Publicado: 2 de septiembre de 2012
20 - 40 | 20 -60 | 20 - 80 | 20 - 100 | 20 - 120 | 20 - 140 | 20 - 160 |
2.587 | 2.986 | 3.712 | 4.526 | 5.026 | 5.226 | 5.796 |
2.072 | 3.09 | 3.906 | 4.606 | 4.956 | 5.49 | 5.826 |
2.136 |3.486 | 3.817 | 4.288 | 5.106 | 5.366 | 5.776 |
2.186 | 3.192 | 3.886 | 4.418 | 4.916 | 5.41 | 5.778 |
2.166 | 2.99 | 3.816 | 4.277 | 4.786 | 5.446 | 5.716 |
∑= 2.229 | ∑= 3.148 | ∑= 3.827 | ∑=4.422 | ∑= 4.958 | ∑= 5.387 | ∑= 5.780 |
Datos obtenidos:
S= f(t)
K = Y1 Y2 –Y3 Y1+Y2-2Y3 --→ec.2 X3= X1 X2---→ec.1
X3= X1 X2
X3=2.229(5.78)
X3= 3.58 Sustituimos en la grafica de la hoja milimétrica para determinar Y3
Y3= 72 cm
Sustituyendo el valor de Y3 en ecuación 2 tenemos que:
K = 40160-(72)^2 40+160-2(72)= 21.71Obtenemos ecuación empírica por medio de mínimos cuadrados, para llegar a
Y= axb + k
x | y | Log x | Log Y-K | LogX LogY | Log x^2 |
2.229 | 18.29 | 0.348 | 1.262 | 0.439 | 0.121 |
3.148 | 38.29 |0.498 | 1.583 | 0.788 | 0.248 |
3.827 | 58.29 | 0.582 | 1.765 | 1.027 | 0.338 |
4.422 | 78.29 | 0.645 | 1.893 | 1.22 | 0.416 |
4.958 | 98.29 | 0.695 | 1.992 | 1.384 | 0.483 |
5.387 | 118.29 |0.731 | 2.072 | 1.514 | 0.534 |
5.78 | 138.29 | 0.761 | 2.14 | 1.628 | 0.579 |
∑= 29.751 | ∑= 528.03 | ∑= 4.26 | ∑= 12.707 | ∑= 8 | ∑= 2.719 |
Sustituimos valores en:
∑Log Y-K = n Log a + b ∑Log X
∑ LogX LogY-K = Log a ∑ Log X + b ∑ Log X2
-12.707 = 7Log a + b 4.26 ec.1 * Log a = 12.707 – 4.26 b Sustituir en ec. 1
8 = Log a 4.26 + b 2.719ec.2 7
8 = 12.707 – 4.26 b (4.26) + 2.719 b
7
8 = 54.13 – 18.147 b + 2.719 b
7
8 = 7.73 – 2.592 b + 2.719b
8 = 7.73 + 0.127 b
B = 0.27 = 2.12
0.127
Sustituyendo b en ec. 2 tenemos que:
Log a = 12.707 – 9.03
7
10Log a = 10(12.707 – 9.03/ 7 )
a = 3.35
El...
2.587 | 2.986 | 3.712 | 4.526 | 5.026 | 5.226 | 5.796 |
2.072 | 3.09 | 3.906 | 4.606 | 4.956 | 5.49 | 5.826 |
2.136 |3.486 | 3.817 | 4.288 | 5.106 | 5.366 | 5.776 |
2.186 | 3.192 | 3.886 | 4.418 | 4.916 | 5.41 | 5.778 |
2.166 | 2.99 | 3.816 | 4.277 | 4.786 | 5.446 | 5.716 |
∑= 2.229 | ∑= 3.148 | ∑= 3.827 | ∑=4.422 | ∑= 4.958 | ∑= 5.387 | ∑= 5.780 |
Datos obtenidos:
S= f(t)
K = Y1 Y2 –Y3 Y1+Y2-2Y3 --→ec.2 X3= X1 X2---→ec.1
X3= X1 X2
X3=2.229(5.78)
X3= 3.58 Sustituimos en la grafica de la hoja milimétrica para determinar Y3
Y3= 72 cm
Sustituyendo el valor de Y3 en ecuación 2 tenemos que:
K = 40160-(72)^2 40+160-2(72)= 21.71Obtenemos ecuación empírica por medio de mínimos cuadrados, para llegar a
Y= axb + k
x | y | Log x | Log Y-K | LogX LogY | Log x^2 |
2.229 | 18.29 | 0.348 | 1.262 | 0.439 | 0.121 |
3.148 | 38.29 |0.498 | 1.583 | 0.788 | 0.248 |
3.827 | 58.29 | 0.582 | 1.765 | 1.027 | 0.338 |
4.422 | 78.29 | 0.645 | 1.893 | 1.22 | 0.416 |
4.958 | 98.29 | 0.695 | 1.992 | 1.384 | 0.483 |
5.387 | 118.29 |0.731 | 2.072 | 1.514 | 0.534 |
5.78 | 138.29 | 0.761 | 2.14 | 1.628 | 0.579 |
∑= 29.751 | ∑= 528.03 | ∑= 4.26 | ∑= 12.707 | ∑= 8 | ∑= 2.719 |
Sustituimos valores en:
∑Log Y-K = n Log a + b ∑Log X
∑ LogX LogY-K = Log a ∑ Log X + b ∑ Log X2
-12.707 = 7Log a + b 4.26 ec.1 * Log a = 12.707 – 4.26 b Sustituir en ec. 1
8 = Log a 4.26 + b 2.719ec.2 7
8 = 12.707 – 4.26 b (4.26) + 2.719 b
7
8 = 54.13 – 18.147 b + 2.719 b
7
8 = 7.73 – 2.592 b + 2.719b
8 = 7.73 + 0.127 b
B = 0.27 = 2.12
0.127
Sustituyendo b en ec. 2 tenemos que:
Log a = 12.707 – 9.03
7
10Log a = 10(12.707 – 9.03/ 7 )
a = 3.35
El...
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