Resumen
El propósito del estudio era determinar la longitud de la altura ,el diámetro
externo, el diámetro interno y la masa de 2 cilindros huecos con dos
calibradores de diferente precision(p1 = 0.05, p2 = 0.02), se logró obtener las
medidas mediante una serie de mediciones, se calculó los errores estimados
y porcentuales y los valores ya requeridos satisfactoriamente.
Análisis de datos.
a) Cilindro1(p=0.05[mm])
Altura(H1)
74,35 + 74,35 + 74,20 + 74,15 + 74,20 +74,20
̅ = 1 ∑𝑛=6
1) Hrep = 𝐻
𝐻𝑖 =
= 74,24166667[mm]
n 𝑖=1
6
Hrep = 74,24[mm]
𝑛=6
2)
𝑑i
0.03709333333
σ n-1 = √∑𝑖=1
=√
=0.08612007121[mm]
𝑛−1
6−1
2
0.08612007121[mm]
σ𝐻̅= 𝛔 n−1
=
= 0.03515837185[mm]
√n
√6
σ𝐻̅= 0.04 [mm]
eH = max[p,σH̅] =max{0.05[mm], 0.04 [mm]}
eH = 0.05 [mm]
𝒆H
0.05[m𝑚]
3) E% = | H𝑟𝑒𝑝| x 100% = | 74.24[m𝑚]| x 100% = 6.7%
H1 = (74.24 ± 0.05)[mm]; 6.7 %
Diámetro externo(D1)
49.60+49.65+49.70+49.65+49.65+49.65
̅ = 1 ∑𝑛=6
1) Hrep = D
D𝑖 =
= 49.65[mm]
n 𝑖=1
6
Drep = 49,65[mm]
2)
2
∑𝑛=6
𝑖=1 𝑑i
σ n-1 = √
𝑛−1
σ n-1 = √(49.65-49.60)
2 +(49.65-49.65)2 +(49.65-49.70)2 +(49.65-49.65)2 +(49.65-49.65)2 +(49.65-49.65)2
6−1
σ n-1 = 0.0316227766 [mm]
σD̅= 𝛔 n−1
√n
=
0.0316227766 [mm]
√6
= 0.01290994449 [mm]
σD̅= 0.01 [mm]
eD = max[p,σD̅] =max{0.05[mm], 0.01 [mm]}
eD = 0.05 [mm]
𝒆D
0.05[m𝑚]
3) E% = | D𝑟𝑒𝑝| x 100% = | 49.65[m𝑚]| x 100% = 0.1%
D1 = (49.65 ± 0.05)[mm]; 0.1 %
Diámetro interno (d1)
40.50 + 40.70 + 40.60 + 40.75 + 40.80 +40.70
̅ = 1 ∑𝑛=6
1) drep = d
= 40.675 [mm]
𝑖=1 d𝑖 =
n
6
drep = 40.68[mm]
2)
2
∑𝑛=6
𝑖=1 𝑑i
σ n-1 = √
𝑛−1
0.05875
6−1
=√
=0.1083974169 [mm]
0.1083974169 [mm]
σd̅= 𝛔 n−1
=
= 0.04425306016[mm]
√n
√6
σd̅= 0.04 [mm]
ed = max[p,σd̅] =max{0.05[mm], 0.04 [mm]}
ed = 0.05 [mm]
3) E% = |
𝒆d
|
d𝑟𝑒𝑝
x 100% = |
0.05[m𝑚]
|
40.68[m𝑚]
x 100% = 1.2 %
d1 = (40.68 ± 0.05) [mm]; 1.2 %
Masa(m1)
mrep = 121.85[g]
em = 0.01[g]
E% = |
𝑒m
|
m𝑟𝑒𝑝
x 100% = |
1 [g]
|
121.85[g]
x 100% = 0.8%
m1 = (121.85 ± 1)[g]; 0.8%
b) Cilindro 2(p=0.02[mm])
Altura(H2)
1)
71.20 + 71.20 + 71.20 + 71.20 + 71.20 +71.10
̅ = 1 ∑𝑛=6
Hrep = 𝐻
𝐻𝑖 =
= 71,18333333[mm]
n 𝑖=1
6
Hrep = 71.18[mm]
2)
2
∑𝑛=6
𝑖=1 𝑑i
8.055555445x10-3
=0.04013864832 [mm]
𝑛−1
6−1
𝛔 n−1
0.04013864832 [mm]
=
= 0.01638653456[mm]
√n
√6
σ n-1 = √
σ𝐻̅=
=√
σ𝐻̅= 0.02 [mm]
eH = max[p,σH̅] =max{0.02[mm], 0.02 [mm]}
eH = 0.02 [mm]
𝒆H
0.02[m𝑚]
3) E% = | H𝑟𝑒𝑝| x 100% = | 71.18[m𝑚]| x 100% = 0.03%
H2 = (71.18 ± 0.02)[mm]; 0.03 %
Diámetro externo(D2)
49.60+49.62+49.70+49.68+49.64+49.68
̅ = 1 ∑𝑛=6
1) Hrep = D
D𝑖 =
= 49.65333333[mm]
n 𝑖=1
6
Drep = 49,65[mm]
2)
2
∑𝑛=6
𝑖=1 𝑑i
σ n-1 = √
𝑛−1
-3
σ n-1 = √733333334x10
6−1
= 0.03932768321[mm]
σ n-1 = 0.03932768321[mm]
0.03932768321[mm]
σD̅= 𝛔 n−1
=
√n
√6
= 0.01605545944 [mm]
̅ = 0.02 [mm]
σD
eD = max[p,σD̅] =max{0.02[mm], 0.02 [mm]}
eD = 0.02 [mm]
𝒆D
0.02[m𝑚]
3) E% = | D𝑟𝑒𝑝| x 100% = | 49.65[m𝑚]| x 100% = 0.04%
D2 = (49.65 ± 0.02)[mm]; 0.04 %
Diámetro interno (d2)
1)
39.90+39.92+39.88+40.04+39.92+39.92
̅ = 1 ∑𝑛=6
drep = d
d𝑖 =
= 39.93 [mm]
n 𝑖=1
6
drep = 39.93[mm]
2)
2
∑𝑛=6
𝑖=1 𝑑i
0.0158
=√
=0.05621387729 [mm]
𝑛−1
6−1
𝛔 n−1
0.05621387729 [mm]
=
= 0.0229492193[mm]
n
√
√6
σ n-1 = √
σd̅=
̅ = 0.02 [mm]
σd
ed = max[p,σd̅] =max{0.02[mm], 0.02 [mm]}
ed = 0.02 [mm]
𝒆d
0.02[m𝑚]
3) E% = | d𝑟𝑒𝑝| x 100% = | 39.93[m𝑚]| x 100% = 0.05 %
d2 = (39.93 ± 0.02) [mm]; 0.05 %
Masa(m2)
mrep = 128.47[g]
em = 1[g]
𝑒m
1 [g]
E% = | m𝑟𝑒𝑝| x 100% = | 128.47[g]| x 100% = 0.8%
m1 = (128.47 ± 1)[g]; 0.8%
