Roger, you're right. I am stupid. Thank you for so kindly for
pointing that out. We all need more "Elmers" like you.


No, you're probably not stupid... maybe ignorant. You can fix that. If
you don't own a copy of the ARRL Handbook, then GET ONE! Its the best
single source of info you can get, and you will find answer to most if not
all of your questions in there.



Ed WB6SAT

Instead of posting look the formula up in your study guide. Learn how to
find your information in the guides and Tec manuals first. Go out and pick
up a copy of the ARRL Handbook and learn how to use it. It has the basic
formulas in it and how to put them to practical use. I had to question how
you got your license also not knowing that formula. I will give you 1 piece
of advice. When you are working with metric DON'T Convert just keep it all
metric.

"Moody1951" wrote in message
...
Roger, you're right. I am stupid. Thank you for so kindly for pointing
that
out. We all need more "Elmers" like you.

Geez...maybe he'll convert this over to U.S. measurement!


"Roger Conroy" wrote in message
...

"Moody1951" wrote in message
...
Please help me out with the math. What is the length of a 5/8s
wavelength
2
meter antenna? I've got something that might be able to be cut to the
freq.

Thanks

Lets assume you want a center frequency of 146 megahertz...

300/146=2.057945m (round it to nearest mm 2.058m)
2.058/8=0.25725m
0.25725x5=1.28625m (round it to nearest mm 1.286m)

Now I have a few questions...

How the ^%$#(*&$ do you get to even own a 2m radio without knowing how to
do
this ELEMENTARY calculation? Do you have a licence? Have you learnt
absolutely nothing? This formula is the first one in the book together
with
Ohm's Law!

73
Roger ZR3RC

On Tue, 6 Apr 2004 09:27:58 +0200, "Roger Conroy"
wrote:

Now I have a few questions...

How the ^%$#(*&$ do you get to even own a 2m radio without knowing how to do
this ELEMENTARY calculation? Do you have a licence? Have you learnt
absolutely nothing? This formula is the first one in the book together with
Ohm's Law!

Perhaps he took the test over a year ago, has only been using a HT,
and is now wanting to get into mobile use?

Let me ask you a few questions:

How far in advance of a turn are you required to use your turn signal?

What is the maximum distance you may enter a bike lane prior to making
a turn?

You may use your hi beams if a car coming towards you is more than
____ feet away, or if a car is in front of you in the same direction
as you, more than _____ feet away.

Gee, you don't know? How the &*^#*$@ did you get your drivers license?

Get the point?

Evan

To reply, remove TheObvious from my e-mail address.

"Evan Platt" wrote in message
...
On Tue, 6 Apr 2004 09:27:58 +0200, "Roger Conroy"
wrote:

Now I have a few questions...

How the ^%$#(*&$ do you get to even own a 2m radio without knowing how to
do
this ELEMENTARY calculation? Do you have a licence? Have you learnt
absolutely nothing? This formula is the first one in the book together
with
Ohm's Law!

Perhaps he took the test over a year ago, has only been using a HT,
and is now wanting to get into mobile use?

Let me ask you a few questions:

How far in advance of a turn are you required to use your turn signal?

What is the maximum distance you may enter a bike lane prior to making
a turn?

You may use your hi beams if a car coming towards you is more than
____ feet away, or if a car is in front of you in the same direction
as you, more than _____ feet away.

Gee, you don't know? How the &*^#*$@ did you get your drivers license?

Get the point?

Evan

To reply, remove TheObvious from my e-mail address.

OK I admit I did come on a bit too strong, I'm sorry.

However, your analogy with drivers licence test questions is a bit off
base...
Not knowing how to convert between frequency and wavelength is much more
fundamental than knowing the distance to dip your lights.
Its rather like being uncertain about the difference between a steering
wheel and a spare wheel.

Anyway this is the last word I'm saying on this topic.

73
Roger ZR3RC

On Thu, 8 Apr 2004 09:39:22 +0200, "Roger Conroy"
wrote:

However, your analogy with drivers licence test questions is a bit off
base...
Not knowing how to convert between frequency and wavelength is much more
fundamental than knowing the distance to dip your lights.
Its rather like being uncertain about the difference between a steering
wheel and a spare wheel.

Well, to you - yes. To me? No. My operations are primarily HT. I have
a few installed mobiles. They work fine with the antennas I've bought
from friends, so I know they're tuned for the frequency I needed - I
went to my friend, told them I need a antenna for xxx mhz, and he gave
me an antenna. But now let's say I have an antenna that I know is WAY
too long for the frequency I want it on - i.e. I'm taking a 42 mhz
antenna I want to cut to work on 220. For the life of me, I couldn't
recall the length of the antenna.

So, for someone who 99% of the time uses their HT, and then decides to
go mobile, the length of a mobile antenna may be akin to how many feet
in front of a intersection they can turn. Prior to this thread, if
someone would have asked me on the street the formula for cutting an
antenna, I would have guessed somewhere around the right number, but
nowhere near exact.

Evan

--
To reply, remove TheObvious from my e-mail address.

On Thu, 8 Apr 2004 09:39:22 +0200, "Roger Conroy"
wrote:

However, your analogy with drivers licence test questions is a bit off
base...
Not knowing how to convert between frequency and wavelength is much more
fundamental than knowing the distance to dip your lights.
Its rather like being uncertain about the difference between a steering
wheel and a spare wheel.

Well, to you - yes. To me? No. My operations are primarily HT. I have
a few installed mobiles. They work fine with the antennas I've bought
from friends, so I know they're tuned for the frequency I needed - I
went to my friend, told them I need a antenna for xxx mhz, and he gave
me an antenna. But now let's say I have an antenna that I know is WAY
too long for the frequency I want it on - i.e. I'm taking a 42 mhz
antenna I want to cut to work on 220. For the life of me, I couldn't
recall the length of the antenna.

So, for someone who 99% of the time uses their HT, and then decides to
go mobile, the length of a mobile antenna may be akin to how many feet
in front of a intersection they can turn. Prior to this thread, if
someone would have asked me on the street the formula for cutting an
antenna, I would have guessed somewhere around the right number, but
nowhere near exact.

Evan

--
To reply, remove TheObvious from my e-mail address.

"Evan Platt" wrote in message
...
On Tue, 6 Apr 2004 09:27:58 +0200, "Roger Conroy"
wrote:

Now I have a few questions...

How the ^%$#(*&$ do you get to even own a 2m radio without knowing how to
do
this ELEMENTARY calculation? Do you have a licence? Have you learnt
absolutely nothing? This formula is the first one in the book together
with
Ohm's Law!

Perhaps he took the test over a year ago, has only been using a HT,
and is now wanting to get into mobile use?

Let me ask you a few questions:

How far in advance of a turn are you required to use your turn signal?

What is the maximum distance you may enter a bike lane prior to making
a turn?

You may use your hi beams if a car coming towards you is more than
____ feet away, or if a car is in front of you in the same direction
as you, more than _____ feet away.

Gee, you don't know? How the &*^#*$@ did you get your drivers license?

Get the point?

Evan

To reply, remove TheObvious from my e-mail address.

OK I admit I did come on a bit too strong, I'm sorry.

However, your analogy with drivers licence test questions is a bit off
base...
Not knowing how to convert between frequency and wavelength is much more
fundamental than knowing the distance to dip your lights.
Its rather like being uncertain about the difference between a steering
wheel and a spare wheel.

Anyway this is the last word I'm saying on this topic.

73
Roger ZR3RC

Damn Roger... It must be really tough being so superior.


"Roger Conroy" wrote in message
...

"Moody1951" wrote in message
...
Please help me out with the math. What is the length of a 5/8s
wavelength
2
meter antenna? I've got something that might be able to be cut to the
freq.

Thanks

Lets assume you want a center frequency of 146 megahertz...

300/146=2.057945m (round it to nearest mm 2.058m)
2.058/8=0.25725m
0.25725x5=1.28625m (round it to nearest mm 1.286m)

Now I have a few questions...

How the ^%$#(*&$ do you get to even own a 2m radio without knowing how to
do
this ELEMENTARY calculation? Do you have a licence? Have you learnt
absolutely nothing? This formula is the first one in the book together
with
Ohm's Law!

73
Roger ZR3RC

Roger, you're right. I am stupid. Thank you for so kindly for pointing that
out. We all need more "Elmers" like you.