In a message dated 96-09-09 23:47:50 EDT, you write:
>
>> This is a word Problem.
>> If I have a 150 Feet of RG-9913, which the charts say should have
>> a loss of 1.2 dB per 100 Feet at 28MHz. And I put 100 Watts of power
>> into the coax, how much power should I expect to see at the other end?
>> Thanks, Jack, KA8D
>----------------------
>Jack:
>
>
Hi Jack,
The overall attenuation is 1.8 dB, since it is 1.2 dB per 100 ft and you have
150 ft of cable (1.5 times 1.2 dB per 100= 1.8 dB)
The overall power loss at 28 MHz would be about 34 watts, you'd have about 66
watts left if SWR was 1:1 at the load.
73 Tom
>From hwardsil@wolfenet.com (Ward Silver) Tue Sep 10 15:20:11 1996
From: hwardsil@wolfenet.com (Ward Silver) (Ward Silver)
Subject: Cable Attenuation Question
Message-ID: <Pine.OSF.3.95.960910071355.17266A-100000@gonzo.wolfenet.com>
> If I have a 150 Feet of RG-9913, which the charts say should have
> a loss of 1.2 dB per 100 Feet at 28MHz. And I put 100 Watts of power
> into the coax, how much power should I expect to see at the other end?
> Thanks, Jack, KA8D
1.2dB loss/100 ft x 150 ft = 1.8dB loss
Pwr out -1.8
10 x log(-------) = -1.8 so Pwr out = Pwr in (antilog (----))
Pwr in 10
= 100 (antilog (-0.18)
= 100 (0.661)
= 66.1W
73, Ward N0AX
>From hwardsil@wolfenet.com (Ward Silver) Tue Sep 10 15:37:46 1996
From: hwardsil@wolfenet.com (Ward Silver) (Ward Silver)
Subject: Cable Attenuation Question
Message-ID: <Pine.OSF.3.95.960910073605.17266G-100000@gonzo.wolfenet.com>
On Mon, 9 Sep 1996, Jeff Singer wrote:
>
> The answer is 75 watts (I'm not so smart; this from a little program on
> QRZ CD).
>
> 73 de Jeff WA2SYN
BZZZZZT! 75 watts is what comes out the other end of 100-feet. Jack has
150-feet for 1.8dB loss.
"I'll take coaxial attenuation for $300, please?"
73, Ward N0AX
>From n5ia@juno.com (Milt Jensen) Tue Sep 10 15:39:32 1996
From: n5ia@juno.com (Milt Jensen) (Milt Jensen)
Subject: To correct or not
References: <199609101350.IAA09773@firefly.prairienet.org>
Message-ID: <19960910.143108.6974.0.N5IA@juno.com>
---Milt Jensen @ ARS N5IA---
---Virden, NM Route Box 176---
---Duncan, AZ -- 85534---
---H: (505) 358-2105 W: (520) 359-2503---
---Reply to n5ia@juno.com---
Comment regarding
On Tue, 10 Sep 1996 08:50:46 -0500 (CDT) w9sz@prairienet.org (Zack Widup)
writes:
>
>
>>
>> Maybe we will have a race between KN5S and N4BO for this contest
>(or
>>maybe KM5S will show up in a few logs)...
>>
>>Hey, don't forget about W6HB !!
>>Bobby
>>WB5B
>>
>
>The best one with my call so far: in the ARRL DX CW CONTEST a few
>years
>ago, someone logged me as W1ESZ :-)
>
>Zack W9SZ
At times because of the way some operators scribble my call sign on their
notepads, I am responded to as a very exotic DX station
with the call sign of n51a (N fifty one Alpha). That is my only claim
to fame.
73 de Milt, Number 5 Intelligent Amateur
>From seay@alaska.net (Del Seay) Tue Sep 10 14:42:01 1996
From: seay@alaska.net (Del Seay) (Del Seay)
Subject: Cable Attenuation Question
References: <199609101339.IAA08494@firefly.prairienet.org>
Message-ID: <32357029.12D8@alaska.net>
Zack Widup wrote:
>
> >
> >> This is a word Problem.
> >> If I have a 150 Feet of RG-9913, which the charts say should have
> >> a loss of 1.2 dB per 100 Feet at 28MHz. And I put 100 Watts of power
> >> into the coax, how much power should I expect to see at the other end?
> >> Thanks, Jack, KA8D
> >----------------------
> >Jack:
> >
> >The answer is 75 watts (I'm not so smart; this from a little program on
> >QRZ CD).
> >
> >73 de Jeff WA2SYN
> >js@li.net
>
> I get 66 watts out for 150 feet from my hand calculator. I get 75 watts
> out for 100 feet of coax.
>
> Is this a new form of CONTEST? :-)
>
> Zack W9SZ
>
> --
Great subject: Showed me my math gets worse as I get older.
I originally told Jack it was 52 watts - miscalculated!!
1.2 dB X 1.50 (150')= 1.8 dB loss.
1.8 dB = .6607, so obviously it would be 66 watts properly terminated
and no loss in connectors.
I had converted to dBm (50 dBm = 100 watts), subtracted 1.8dB and
came up with 52 watts. Gawd I'm good!
73 de KL7HF
>From weinfurtner@ouvaxa.cats.ohiou.edu (Greg Weinfurtner) Tue Sep 10 14:46:00
>1996
From: weinfurtner@ouvaxa.cats.ohiou.edu (Greg Weinfurtner) (Greg Weinfurtner)
Subject: Cable attenuation
Message-ID: <v03007802ae5b1e2ee04a@[132.235.72.11]>
Gang,
This is a good problem! Here is the formula for db
db= 10 log (power IN/OUT Ratio)
A. His total loss in 150' is 1.8db (That is 1.2db per 100 feet)
B. -1.8db= 10 log (x/100)
C. -.18db= log (x/100)
D. 10 ^ -.18= .661 (^ means exponential or "to the power of")
E. 100 * .661= 66.1 WATTS
F. Plug it back into the formula and voila!, seems to check OK!
CU all this month in the contests! 73 de
****************************************************************************
* Greg Weinfurtner AEE BSS *
* NN N SSSSSSS 8888888 OOOOOOO Electronic Design Splst *
* N N N S 8 8 O O Ohio University Athens *
* N N N SSSSSSS 8888888 O O GO BOBCATS! *
* N N N S 8 8 O O *
* N NN SSSSSSS 8888888 OOOOOOO *
* Can thou send forth lightnings *
* Amateur Radio that they may go and say unto *
* DXCC WAS EM89 thee, 'Here we are'? Job 38:35 *
* weinfurtner@ouvaxa.cats.ohiou.edu *
* http://ouvaxa.cats.ohiou.edu/~weinfurtner *
****************************************************************************
>From wa2syn@li.net (Jeff Singer) Tue Sep 10 15:49:51 1996
From: wa2syn@li.net (Jeff Singer) (Jeff Singer)
Subject: Cable Attenuation Question
Message-ID: <Pine.SUN.3.95.960910104451.12150D-100000@linet01>
OK, Zack, and everyone else who brought the error to my attention: ur
absolutely right. Thought he was using 100' instead of 150. That's what I
get for reading my mail after 11pm (and I DID say I wasn't so smart)!
73 de Jeff WA2SYN
js @li.net
------------------------------------------
On Tue, 10 Sep 1996, Zack Widup wrote:
>
>
> >> If I have a 150 Feet of RG-9913, which the charts say should have
> >> a loss of 1.2 dB per 100 Feet at 28MHz. And I put 100 Watts of power
> >> into the coax, how much power should I expect to see at the other end?
> >> Thanks, Jack, KA8D
> >----------------------
> >Jack:
> >
> >The answer is 75 watts (I'm not so smart; this from a little program on
> >QRZ CD).
> >
> >73 de Jeff WA2SYN
> >js@li.net
>
> I get 66 watts out for 150 feet from my hand calculator. I get 75 watts
> out for 100 feet of coax.
>
> Is this a new form of CONTEST? :-)
>
> Zack W9SZ
>
> --
>
>
|