Respuesta :
Answer:
The total time required is [tex]T_{total}= 1601 sec[/tex]
Explanation:
From the question we are told that the
The thickness of the aluminium plate is [tex]t = 1.75 \ in[/tex]
The diameter of the holes is [tex]d = \frac{3}{4} \ in[/tex]
The number holes is n = 100
The pattern they arranged is = 10 by 10 matrix
The distance between the adjacent holes is [tex]D = 1.5 \ in[/tex]
The cutting speed [tex]v = 300 \ ft/min =300 *12 = 3600 \ in /min[/tex]
The penetration feed is [tex]P = 0.015 \ in / rev[/tex]
The traverse rate is [tex]T_r = 15.0 \ in /min[/tex]
The distance of the x-y moves [tex]L = 0.50 \ in[/tex]
The rate at which the drill id retracted is [tex]R = 2P[/tex]
The point angle of the drill is [tex]\theta = 100^o[/tex]
Generally the cutting speed is mathematically represented as
[tex]v = \pi d N[/tex]
Where N is the rotational speed
Substituting values
[tex]3600 = \pi * \frac{3}{4} N[/tex]
=> [tex]N = \frac{3600}{0.75 * 3.142 }[/tex]
[tex]N = 1527.68 rev /min = \frac{1527.68}{60} = 25.46 \ rev /sec[/tex]
Now the feed rate is mathematically represented as
[tex]Feed \ Rate [F] = P * N[/tex]
Substituting values
[tex]F = 0.015 * 25.46[/tex]
[tex]= 0.3819 \ in[/tex]
Next is to obtain the approach distance
The approach distance is mathematically represented as
[tex]a = 0.5 * d * tan (90 - \frac{\theta }{2} )[/tex]
substituting values
[tex]a = 0.5 * 0.75 * tan ({90 - \frac{100}{2} })[/tex]
[tex]a = 2.5646 \ in[/tex]
also we are told that the drill x-y moves is 0.5 in above the work surface
Then the when retracting the drill the total distance moves is
Hence the total drill distance is [tex]w = 0.5 + 1.75 + 0.3146[/tex]
[tex]= 2.5646 \ in[/tex]
The time required to drill on hole is
[tex]t_1 = \frac{w }{F }[/tex]
Substituting values
[tex]t_1 = \frac{2.56446}{0.3819}[/tex]
[tex]= 6.71 sce[/tex]
The retracting rate of the drill would be
[tex]K = 2P N[/tex]
[tex]R = 0.015 * 2 * 25.46[/tex]
[tex]= 0.7638[/tex]
The time required to move the drill out of one hole is
[tex]t_ 2 = \frac{w}{K}[/tex]
Substituting value
[tex]t_2 = \frac{2.5646}{0.7638}[/tex]
[tex]= 3.3576 \ sec[/tex]
The time required to move from one hole to another is
[tex]t_3 = \frac{D}{T_r}[/tex]
[tex]= \frac{1.5}{0.25}[/tex]
[tex]= 6 \ sec[/tex]
The time it would take to perform the drilling of 100 holes is mathematically represented as
[tex]t_4 = 100 *t_1[/tex]
[tex]= 6.71 *100[/tex]
[tex]= 671 \ sec[/tex]
Time required for retracting the drill from 100 holes is mathematically represented as
[tex]t_5 = 100 * t_2[/tex]
[tex]t_5 = 3.3576 *100[/tex]
[tex]= 336 sec[/tex]
The time taken to move through the 100 holes is
[tex]t_6 = number of travers time * t_3[/tex]
The number of trans verse time is = 99
[tex]t_6 = 99 * 6[/tex]
[tex]= 594 \ sec[/tex]
The time required the drilling task from the beginning of the first hole to the hole is
[tex]T_{total } = t_4 + t_5 + t_6[/tex]
[tex]= 671 + 336 + 594[/tex]
[tex]T_{total}= 1601 sec[/tex]
