Physics, 23.03.2020 18:33, nila49

A face milling operation is used to machine 6.0 mm from the top surface of a rectangular piece of aluminum 300 mm long by 125 mm wide in a single pass. The cutter follows a path that is centered over the workpiece. It has four teeth and is 150 mm in diameter. Cutting speed

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answered: hbstanley3

Complete Question

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The machining time is T_m = 0.866minutes

The maximum metal removal rate is $R_{mr} =4812.75mm^3 /sec$

Explanation:

From the question we are told that

The length of the work piece is L =300mm

The width of the work piece is W = 125mm

The cut Depth is D = 6mm

The chip load is f = 0.27 mm/tooth

The number of teeth is n = 4

The cutting speed is v = 2.8m/s = 2800mm/s

Now in order to obtain the actual machining time to make the pass across the surface

We first obtain the speed of rotation of the machine

The cutting velocity is mathematically represented as

$v = \pi d N$

Where d is the diameter given as 150mm

N is the speed of rotation

Making N the subject of the formula

$N =\frac{v}{\pi *d}$

$= \frac{2800}{\pi *150 }$

= 5.941rev/sec

Next we need to obtain the feed rate this is mathematically represented as

f_r = N nf

Substituting the values accordingly

= 5.94 * 4 * 0.27

= 6.417 mm/s

The displacement of the milling operation is mathematically represented as

$A = 0.5(d - \sqrt{d^2 - W^2} )$

$= 0.5(150 - \sqrt{150^2 - 125^2} )$

= 0.5 * (150 - 82.9)

= 33.5mm

The machining time is mathematically represented as

$T_m = \frac{(L + A)}{f_r}$

$= \frac{(300 +33.5)}{6.417}$

$=\frac{51.97}{60} = 0.866min$

The maximum metal removal rate is mathematically represented as

$R_{mr} = W D f_r$

Substituting values accordingly

= 125 * 6* 6.417

= 4812.75 mm^3/sec

A face milling operation is used to machine 6.0 mm from the top surface of a rectangular piece of al

answered: Guest

the results is a decrease in air temperature with an increase of altitude.

explanation:

answered: Guest

we have been given a quadratic function $f(x)=(x-5)^{2} +2$ and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is $[2,\infty)$ .

if we restrict the domain of this function to either $(-\infty,5]$ or $[5,\infty)$ , it will become one to one function.

let us know find its inverse.

$y=(x-5)^{2}+2$

upon interchanging x and y, we get:

$x=(y-5)^{2}+2$

let us now solve this function for y.

$(y-5)^{2}=x-2\\ y-5=\pm \sqrt{x-2}\\ y=5\pm \sqrt{x-2}\\$

hence, the inverse function would be $f^{-1}(x)=5+\sqrt{x-2}$ if we restrict the domain of original function to $[5,\infty)$ and the inverse function would be $f^{-1}(x)=5-\sqrt{x-2}$ if we restrict the domain to $(-\infty,5]$ .

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A face milling operation is used to machine 6.0 mm from the top surface of a rectangular piece of al...