# Thread: is this complete ?

1. ## is this complete ?

using the equation VT(power n) = C

cuttin speed tool life
180m/min 30
250m/min 12

v1=180
t1=30
v2=250
t2=12

logv1+logt1=log c
logv2+logt2=log c

log180+nlog30
log250+nlog12

2.255+n1.477
-2.397+n1.079
=-0.142+n-0.602=0

n=0.602 = 0.142

0.142
0.602
=0.235

180+0.235x1.477=180.3470

is this the end of the equation?

2. sorry missed out a bit of infomation n=0.13

3. Originally Posted by jenko
sorry missed out a bit of infomation n=0.13
But as I showed in my last post, n = 0.13 isn't going to work in your equation. If $VT^n = C$ is correct for both sets of data then your problem is inconsistent.

Are you giving us the full problem? If so, then what are you trying to find?

-Dan

4. full equation:

a test was carried out on the centre lathe and the following results obtained:

Cutting speed (V) Tool life (T)
180m/min 30min
250m/min 12min

using the equation VT(small n at the top of T) = C

5. Originally Posted by jenko
full equation:

a test was carried out on the centre lathe and the following results obtained:

Cutting speed (V) Tool life (T)
180m/min 30min
250m/min 12min

using the equation VT(small n at the top of T) = C
So you are trying to find n and C?

-Dan

6. i think thats what its asking :S i am realy unsure i do not have any other infomation on this question

7. Originally Posted by jenko
full equation:

a test was carried out on the centre lathe and the following results obtained:

Cutting speed (V) Tool life (T)
180m/min 30min
250m/min 12min

using the equation VT(small n at the top of T) = C
Okay, start to finish, finding n and C. Here's how I would do this:
First
$VT^n = C$

So
$V_1T_1^n = V_2T_2^n$

$180 \cdot 30^n = 250 \cdot 12^n$

$\frac{30^n}{12^n} = \frac{250}{180}$

$\left ( \frac{30}{12} \right )^n = \frac{25}{18}$

$\left ( \frac{5}{2} \right )^n = \frac{25}{18}$

$log \left [ \left ( \frac{5}{2} \right )^n \right ] = log \left ( \frac{25}{18} \right )$

$n~log \left ( \frac{5}{2} \right ) = log \left ( \frac{25}{18} \right )$

$n = \frac{log \left ( \frac{25}{18} \right )}{log \left ( \frac{5}{2} \right )} = 0.358515$
(which means I screwed up in my earlier post. )

Now we may take either set of data to find C:
180m/min 30min
$180 \cdot 30^{0.358515} = 609.316$

-Dan