Re: [Elecraft] OT: math question
And while we're on the subject... the correct unit is dB, not dBm. Remember that dB is a relative term. In the case of amplifier gain, this is relative. The unit dBm is an absolute unit relative to 1 milliWatt. When the amplifier is finally built, it will have 50 dBm of output power (assuming it is driven with enough input signal) and will have 13 dB of gain. Al. - Original Message - From: S55M [EMAIL PROTECTED] To: [EMAIL PROTECTED]; [EMAIL PROTECTED]; elecraft@mailman.qth.net Sent: Wednesday, March 30, 2005 12:41 PM Subject: Re: [Elecraft] OT: math question dBm's Just remember this (no maths:))) : Every 3dB increase is power multiplied by 2 times (1W to 2W is 30dBm to 33dBm) So 1W+3dB (1x2=2) is 2W+3dB (2x2=4)4W+3dB(4x2=8)8W and so on. Every 10dB increase is power multiplied by 10 (1W to 10W is 30dBm to 40dBm 100W to 1KW is 50 dBm to 60 dBm the easy one HI) And dont care abt digits on calcullator (try to see 0,5dB on Your S-meter or ask someone on the band to change the power from 100W to 50W or from 10W to 5W (3dB) You will see practically no difference. Why dBm? Because dBm is power related to 1mW=0dBm . S55M-Adi ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
RE: [Elecraft] OT: math question
dB = 10 log (P2/P1) For your example of 5 watts to 100 watts, is an increase of 13.01029996 dB - pardon the rounding but that is all the digits my calculator shows :). 73, Don W3FPR -Original Message- How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. K3UJ ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
dBm's Just remember this (no maths:))) : Every 3dB increase is power multiplied by 2 times (1W to 2W is 30dBm to 33dBm) So 1W+3dB (1x2=2) is 2W+3dB (2x2=4)4W+3dB(4x2=8)8W and so on. Every 10dB increase is power multiplied by 10 (1W to 10W is 30dBm to 40dBm 100W to 1KW is 50 dBm to 60 dBm the easy one HI) And dont care abt digits on calcullator (try to see 0,5dB on Your S-meter or ask someone on the band to change the power from 100W to 50W or from 10W to 5W (3dB) You will see practically no difference. Why dBm? Because dBm is power related to 1mW=0dBm . S55M-Adi - Original Message - From: W3FPR - Don Wilhelm [EMAIL PROTECTED] To: [EMAIL PROTECTED]; elecraft@mailman.qth.net Sent: Wednesday, March 30, 2005 8:23 PM Subject: RE: [Elecraft] OT: math question dB = 10 log (P2/P1) For your example of 5 watts to 100 watts, is an increase of 13.01029996 dB - pardon the rounding but that is all the digits my calculator shows :). 73, Don W3FPR -Original Message- How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. K3UJ ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
In a message dated 3/30/2005 10:02:44 AM Pacific Standard Time, [EMAIL PROTECTED] writes: How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. The formula is: dB = 10 log (P2 / P1) P2 / P1 = 100 / 5 = 20 log of 20 = 1.30103 10 times it = 13.0103 That's your answer: 13.0103 dB Geoff, K6TFZ ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
Im with Adi on this one..i.e NO MATH!. Although I'm 100% certain Don is correct, my fingers came with *about* a 12.6 dB increase. (I never did understand those loggers) Tom WB2QDG K2 1103 -- Original message -- From: S55M [EMAIL PROTECTED] dBm's Just remember this (no maths:))) : Every 3dB increase is power multiplied by 2 times (1W to 2W is 30dBm to 33dBm) So 1W+3dB (1x2=2) is 2W+3dB (2x2=4)4W+3dB(4x2=8)8W and so on. Every 10dB increase is power multiplied by 10 (1W to 10W is 30dBm to 40dBm 100W to 1KW is 50 dBm to 60 dBm the easy one HI) And dont care abt digits on calcullator (try to see 0,5dB on Your S-meter or ask someone on the band to change the power from 100W to 50W or from 10W to 5W (3dB) You will see practically no difference. Why dBm? Because dBm is power related to 1mW=0dBm . S55M-Adi - Original Message - From: W3FPR - Don Wilhelm [EMAIL PROTECTED] To: [EMAIL PROTECTED]; elecraft@mailman.qth.net Sent: Wednesday, March 30, 2005 8:23 PM Subject: RE: [Elecraft] OT: math question dB = 10 log (P2/P1) For your example of 5 watts to 100 watts, is an increase of 13.01029996 dB - pardon the rounding but that is all the digits my calculator shows :). 73, Don W3FPR -Original Message- How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. K3UJ ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
At 12:01 PM 3/30/05, you wrote: How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. 10 log(P2/P1) = 10 log(100/5) = 10 log(20) = 10 (1.30103) = 13.0103dB An easy 'rule of thumb' way to get a good idea of such levels is to remember that: a power INcrease of 2 = approx. +3dB change a power DEcrease of 2 = approx. -3dB change a power INcrease of 10 = +10dB a power DEcrease of 10 = -10dB So, to go from 5W to 10W output, you have 3dB GAIN to go from 10W to 100W, you have 10dB gain Since dB's can be added, going from 5W to 100W = 3db + 10 dB = 13dB Similarly, if you went from 2W to 800W... From 2W to 4W (first doubling of power) = +3dB From 4W to 8W (2nd doubling of power) = +3dB From 8W to 800W (increase of 10X) = +10dB --- Total power increase in dB = +16dB 73, Tom Hammond N0SS ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
RE: [Elecraft] OT: math question
I ran into a brain-teaser once, that had me baffled for hours (I won't say how many). Finally, I called my sister, the Rocket Scientist (actually, we're not _Rocket_ scientists - That's a quote from her husband) and asked her. She and Paul spent a few minutes on the problem, figured it out, but it was such a trivial answer they threw it away. Since I'd spent so long on the Question, I gave her about a week and then I called back. Luckily they'd not taken out the trash, so she easily found their notes and read them to me. I won't spoil it for you 'crafters. Dan / WG4S / K2 #2456 Problem: An alley has a 20-foot ladder across it, in the left lower corner and leaning on the right wall. There also is a 30-foot ladder, from the right lower corner leaning on the left wall. The laders cross at 10 feet high. How wide is the alley? Usual conventions: alley is all right angles, ladders have no thickness nor bends, blah blah blah... This should make some juices flow out there. Just simple geometry, or is it? PS: Iteration is cheating. Sorry Issac. snip the responses would be far more interesting than merely looking up the answer /snip ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
Yes math is beautifull (callculating needed space for 3d antennas HI. But google can spare some time and paper. :) http://home.att.net/~numericana/answer/algebra.htm#quartic - Original Message - From: Dan Barker [EMAIL PROTECTED] To: [EMAIL PROTECTED]; Elecraft elecraft@mailman.qth.net Sent: Wednesday, March 30, 2005 10:22 PM Subject: RE: [Elecraft] OT: math question I ran into a brain-teaser once, that had me baffled for hours (I won't say how many). Finally, I called my sister, the Rocket Scientist (actually, we're not _Rocket_ scientists - That's a quote from her husband) and asked her. She and Paul spent a few minutes on the problem, figured it out, but it was such a trivial answer they threw it away. Since I'd spent so long on the Question, I gave her about a week and then I called back. Luckily they'd not taken out the trash, so she easily found their notes and read them to me. I won't spoil it for you 'crafters. Dan / WG4S / K2 #2456 Problem: An alley has a 20-foot ladder across it, in the left lower corner and leaning on the right wall. There also is a 30-foot ladder, from the right lower corner leaning on the left wall. The laders cross at 10 feet high. How wide is the alley? Usual conventions: alley is all right angles, ladders have no thickness nor bends, blah blah blah... This should make some juices flow out there. Just simple geometry, or is it? PS: Iteration is cheating. Sorry Issac. snip the responses would be far more interesting than merely looking up the answer /snip ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
I always like to ask... Is this a trick question? The way the problem is worded (and if I understand your geometry correctly) implies an infinite number of solutions as long as the 20 foot ladder can span the alley. Therefore, the width is anything less than or equal to 20 feet. Al. - Original Message - From: Dan Barker [EMAIL PROTECTED] To: [EMAIL PROTECTED]; Elecraft elecraft@mailman.qth.net Sent: Wednesday, March 30, 2005 2:22 PM Subject: RE: [Elecraft] OT: math question I ran into a brain-teaser once, that had me baffled for hours (I won't say how many). Finally, I called my sister, the Rocket Scientist (actually, we're not _Rocket_ scientists - That's a quote from her husband) and asked her. She and Paul spent a few minutes on the problem, figured it out, but it was such a trivial answer they threw it away. Since I'd spent so long on the Question, I gave her about a week and then I called back. Luckily they'd not taken out the trash, so she easily found their notes and read them to me. I won't spoil it for you 'crafters. Dan / WG4S / K2 #2456 Problem: An alley has a 20-foot ladder across it, in the left lower corner and leaning on the right wall. There also is a 30-foot ladder, from the right lower corner leaning on the left wall. The laders cross at 10 feet high. How wide is the alley? Usual conventions: alley is all right angles, ladders have no thickness nor bends, blah blah blah... This should make some juices flow out there. Just simple geometry, or is it? PS: Iteration is cheating. Sorry Issac. snip the responses would be far more interesting than merely looking up the answer /snip ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
RE: [Elecraft] OT: math question
Hmmm ... never seen square root as O with dot-dot on top. Jamie WB4YDL -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of S55M Sent: Wednesday, March 30, 2005 2:47 PM To: Elecraft Subject: Re: [Elecraft] OT: math question Yes math is beautifull (callculating needed space for 3d antennas HI. But google can spare some time and paper. :) http://home.att.net/~numericana/answer/algebra.htm#quartic - Original Message - From: Dan Barker [EMAIL PROTECTED] To: [EMAIL PROTECTED]; Elecraft elecraft@mailman.qth.net Sent: Wednesday, March 30, 2005 10:22 PM Subject: RE: [Elecraft] OT: math question I ran into a brain-teaser once, that had me baffled for hours (I won't say how many). Finally, I called my sister, the Rocket Scientist (actually, we're not _Rocket_ scientists - That's a quote from her husband) and asked her. She and Paul spent a few minutes on the problem, figured it out, but it was such a trivial answer they threw it away. Since I'd spent so long on the Question, I gave her about a week and then I called back. Luckily they'd not taken out the trash, so she easily found their notes and read them to me. I won't spoil it for you 'crafters. Dan / WG4S / K2 #2456 Problem: An alley has a 20-foot ladder across it, in the left lower corner and leaning on the right wall. There also is a 30-foot ladder, from the right lower corner leaning on the left wall. The laders cross at 10 feet high. How wide is the alley? Usual conventions: alley is all right angles, ladders have no thickness nor bends, blah blah blah... This should make some juices flow out there. Just simple geometry, or is it? PS: Iteration is cheating. Sorry Issac. snip the responses would be far more interesting than merely looking up the answer /snip ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
James C. Hall, MD wrote: Hmmm ... never seen square root as O with dot-dot on top. Right, they should have printed it as dah-dah-dah-dit ;-P Sorry, couldn't resist this one 8-) B73, Andrea. -- Homepage: http://andrea.borgia.bo.it /Amateur radio: IZ4FHT ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
RE: [Elecraft] OT: math question
I like simple, 10Db gain will go from 5 to 50Watts; 3Db more will double 50 to 100 Watts. Approximately 13 Db -- 73 Chuck AA8VS www.aa8vs.org/aa8vs FP #113 MI-QRP #1212 SOC #445 Firebird #2117 TSARC #3952 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of [EMAIL PROTECTED] Sent: Wednesday, March 30, 2005 1:02 PM To: elecraft@mailman.qth.net Subject: [Elecraft] OT: math question How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. K3UJ ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
I'll go with an alley of zero width. best wishes, dave belsley, w1euy On Mar 30, 2005, at 3:22 PM, Dan Barker wrote: I ran into a brain-teaser once, that had me baffled for hours (I won't say how many). Finally, I called my sister, the Rocket Scientist (actually, we're not _Rocket_ scientists - That's a quote from her husband) and asked her. She and Paul spent a few minutes on the problem, figured it out, but it was such a trivial answer they threw it away. Since I'd spent so long on the Question, I gave her about a week and then I called back. Luckily they'd not taken out the trash, so she easily found their notes and read them to me. I won't spoil it for you 'crafters. Dan / WG4S / K2 #2456 Problem: An alley has a 20-foot ladder across it, in the left lower corner and leaning on the right wall. There also is a 30-foot ladder, from the right lower corner leaning on the left wall. The laders cross at 10 feet high. How wide is the alley? Usual conventions: alley is all right angles, ladders have no thickness nor bends, blah blah blah... This should make some juices flow out there. Just simple geometry, or is it? PS: Iteration is cheating. Sorry Issac. snip the responses would be far more interesting than merely looking up the answer /snip ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
That can't be right. Drawing it (roughly) on paper shows it to be around 13 feet wide, give or take a foot or so. But coming up with a precise equation takes more math skills than high school left me with... Bill K3UJ In a message dated 3/30/2005 4:51:20 P.M. Eastern Standard Time, [EMAIL PROTECTED] writes: I'll go with an alley of zero width. best wishes, dave belsley, w1euy ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
Hi Frank: Thanks pal... guess I got a bit ahead of myself there! Of course, you're completely correct! Should have been 26db, instead of 16dB! 73, Tom At 03:54 PM 3/30/05, you wrote: Hi Tom, Just a small correction to your 2W to 800W example: From 2W to 4W (first doubling of power) = +3dB From 4W to 8W (2nd doubling of power) = +3dB From 8W to 80W (increase of 10X) = +10dB From 80W to 800W (increase of 10X) = +10dB Total = 3+3+10+10 = 26db Best 73, Frank - W6NEK - Original Message - From: Tom Hammond [EMAIL PROTECTED] To: [EMAIL PROTECTED]; elecraft@mailman.qth.net Sent: Wednesday, March 30, 2005 11:01 AM Subject: Re: [Elecraft] OT: math question At 12:01 PM 3/30/05, you wrote: How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. 10 log(P2/P1) = 10 log(100/5) = 10 log(20) = 10 (1.30103) = 13.0103dB An easy 'rule of thumb' way to get a good idea of such levels is to remember that: a power INcrease of 2 = approx. +3dB change a power DEcrease of 2 = approx. -3dB change a power INcrease of 10 = +10dB a power DEcrease of 10 = -10dB So, to go from 5W to 10W output, you have 3dB GAIN to go from 10W to 100W, you have 10dB gain Since dB's can be added, going from 5W to 100W = 3db + 10 dB = 13dB Similarly, if you went from 2W to 800W... From 2W to 4W (first doubling of power) = +3dB From 4W to 8W (2nd doubling of power) = +3dB From 8W to 800W (increase of 10X) = +10dB --- Total power increase in dB = +16dB 73, Tom Hammond N0SS ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
For convenience, Mini-Circuits has a nice printable conversion table at http://www.mini-circuits.com/dg03-110.pdf Larry N8LP Tom Hammond wrote: At 12:01 PM 3/30/05, you wrote: How would one calculate the amplifier gain (in dB) required to go from one power level to another? For example from 5W to 100W. 10 log(P2/P1) = 10 log(100/5) = 10 log(20) = 10 (1.30103) = 13.0103dB An easy 'rule of thumb' way to get a good idea of such levels is to remember that: a power INcrease of 2 = approx. +3dB change a power DEcrease of 2 = approx. -3dB change a power INcrease of 10 = +10dB a power DEcrease of 10 = -10dB So, to go from 5W to 10W output, you have 3dB GAIN to go from 10W to 100W, you have 10dB gain Since dB's can be added, going from 5W to 100W = 3db + 10 dB = 13dB Similarly, if you went from 2W to 800W... From 2W to 4W (first doubling of power) = +3dB From 4W to 8W (2nd doubling of power) = +3dB From 8W to 800W (increase of 10X) = +10dB --- Total power increase in dB = +16dB 73, Tom Hammond N0SS ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
On Wed, 30 Mar 2005, S55M wrote: Yes math is beautifull (callculating needed space for 3d antennas HI. But google can spare some time and paper. :) http://home.att.net/~numericana/answer/algebra.htm#quartic ANd now I know why my daughters always hated math, and why I never passed the test to get recruited by various federal agencies. I(we) are dumb. 73 Thom ___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
Re: [Elecraft] OT: math question
Well, Bill, my first answer of zero feet width is, in fact, a correct answer, albeit rather degenerate. It was an answer that came immediately to mind. However, there is a nondegenerate answer, which is approximately 12.3119 feet. best wishes, dave belsley, w1euy That can't be right. Drawing it (roughly) on paper shows it to be around 13 feet wide, give or take a foot or so. But coming up with a precise equation takes more math skills than high school left me with... Bill K3UJ In a message dated 3/30/2005 4:51:20 P.M. Eastern Standard Time, [EMAIL PROTECTED] writes: I'll go with an alley of zero width. best wishes, dave belsley, w1euy___ Elecraft mailing list Post to: Elecraft@mailman.qth.net You must be a subscriber to post to the list. Subscriber Info (Addr. Change, sub, unsub etc.): http://mailman.qth.net/mailman/listinfo/elecraft Help: http://mailman.qth.net/subscribers.htm Elecraft web page: http://www.elecraft.com
