Recent content by SLiM6y

HAHA... That's cool berkman. Well, my plan isn't to be a pirate radio station - but a fully legit one. I have now under a month to get my application in! It will be very interesting either way. I appreciate your knowledge on this subject. And even at the least, I plan on incorporating...

It appears the law is quite simple... I have till February 6 to apply for one of three frequencies (cost of application AUD$300) then from there I can get a license from $250 (to a maximum of $2000) depending on residential area. As I live quite rural I think I should be able to get the $250...

Thanks again berkman... 10km is exactly how far I'd like to try and reach that covers my two main townships. I have written to the ACMA (Australian Communications and Media Authority) so I will hear back from them soon. The antennae would probably be about 4 - 6m up on my shed rooftop. If...

Thanks you two... Firstly, I am in Australia and very aware of the laws - I release it is possible I could be some sort of Pirate Radio - but, I believe there is a lower frequency that is capable of being used without license (I am still checking into that). The system I have looked at has...

I have a simple question - but I somehow doubt there is a simple answer. 1) I would like to start a small, non-commercial, home transmitted FM Radio Station - mostly for fun and as a hobby. I have seen an 18W transmitter for a reasonable price. I live on relatively flat land in the country...

But then in that case... that would mean EVERY grandfather clock has a pendulum with a cable length of 0.25 m - does that sound correct? Because the mass doesn't change the period, and the height it is dropped doesn't change the period so then only the length of the cable will change the...

DOH - I get it!

OK - I have looked through it and I can't get the same answer as the book... The question is Calculate the length of cable required to give a clock (grandfather clock) a frequency of 1.0 Hz. So T = 2 pi * sq rt L / g They work the answer through as L = g / T 4 pi^2 = 9.8 / 4 pi^2 = 0.25 m...

Thanks for the reply... I got the same as you the first and the second time, but the book got something different...

Hello, I have a simple equation (maybe simple for some) that I can't understand how to get from one point to the next when re-arranging. If you can help by giving an answer in english I would be thankful... The equation T = 2 pi square root L / g (sorry couldn't find the square root...