Engineering Mechanics of your Landscape Lighting System
This article will be a breath of fresh air for some of you and a gut wrenching process for others. Without getting too involved within the process, I will say this particular piece of the puzzle is designed for the person that thinks with the left side of your brain. Their many studies and books written about the left side and right side of our brains. Many people enjoy a "good" living dealing with issues relating to the left and right side of our brains. This could be the start of another book relating to that subject; however, I am not going there. Too keep this to the simplest form I can, I will say our brain is divided into two sides. We have the left side and the right side.
The left side of the brain functions by way of using logic, facts, knowledge, math and science and is detailed oriented. If you fall into this category, you will enjoy this article. Meanwhile, the right side of the brain functions by using feelings, images, symbols, sees the "big picture", is very imaginative and believes things are possible by taking risks. If you fall into this category, I hope your partner is a left-brain person. This article will be dull and boring for you even though it is an extremely important process for you to achieve your "big picture" on your home to create the "WOW" effect you want.
My wife and I are a good team because she is a professional photographer, thus a right brain person. I am a left-brain person. We compliment each other and guess who takes pictures of my landscape-lighting installations. A good starting subject for this article is "voltage drop". Voltage Drop is created by internal resistance to the flow of electricity through electric wire or cable. It is the difference between the voltages at the beginning of a cable run compared to the voltage at the end of a run. We need to compensate for the Voltage Drop in order to keep the same voltage to all fixtures. If we do not, the end fixtures will have lamps with low voltage and the results will be a decrease in brightness. This is why we can see the yellow looking lights at the end of the runs.
The formula for calculating Voltage Drop is: Total Watts x Length of Run / Cable Size Constant = VOLTAGE DROP. You are now asking yourself, " What is cable size constant". That is a predetermined amount and I will not show the chart for the different size cables. Instead, I will share the constants that you will use for your system. Use either a 10/2-gauge wire, which has a constant of 11920, or a 12/2-gauge wire, which has a constant of 7500.
With this information and the constants, lets figure a "run for fun". This example shows a run for 80 feet and on the run, there are 4 fixtures that have a 20-watt lamp in each. We have a total of 80 watts on a run of 80 feet and 12/2 gauge wire is used for the run. 80-watts x 80 feet =6400, 6400/7500 (constant) = .85 voltage drop. Now subtract .85 (voltage drop) from 12 (amount of volts from source) and we have a resulting voltage reading of 11.15 volts. I like to keep lamps in a range of 10.5 volts to 11.7 volts for peak performance; therefore, the reading of 11.15 volts fits very will within the range. The main thing to remember is to choose the lamp you want and need to do the job you want. Once you decide on the lamp, engineer the system to supply the proper voltage to the lamp.
A rule of thumb on sizing wire for runs is to always go bigger if there is some doubt. The 10/2 gauge wire expense is just slightly higher than the 12/2 gauge wire; however, the peace of mind is well worth the difference. I have included a flowchart to use as an estimate on which gauge wire to use on which run. This does not replace the voltage drop formula and measuring every lead to every fixture using a multi meter to get the proper reading. For runs over 100 feet I recommend using the 12/2 gauge cable just because it can handle more with less voltage drop. QUICK WIRE SIZING GUIDE: Use the guide to select wire sizes for each run - "Are there more than 100 watts lamp load on the run"? If your answer is, yes then you Use #10-2 / If your answer is no then the next question is "Is the run longer than 100 feet? If your answer is no you Use #12-2 and if its Yes then you Use #10-2.
WIRE AMP RATINGS: Wire Size #12/2 100w/8.3A is recommended and the maximum is 192w/16A. Wire Size #10/2 140w/12.0A is recommended and the maximum is 288w/24A. You will notice the low chart gives the wire size with recommended and maximum figures. It gives reference to watts and amps. Remember the formula: (Amps x volts = watts Or Watts/volts = amps) you will be fine on amps using these guidelines. I keep a maximum of 110 watts per run on 12/2 cable and 150 watts on 10/2 cable. I know the system is not working hard and I feel that it will produce an efficient operating system.
Roger Neumann has extensive background in aviation. He started flying when he was 15 years old and his parents had to drive him to the airport because he was too young to drive. His flying career started then and he finished his career with the airlines until the 60 years of age forced him out of the airlines.
During his career, he flew Army helicopters in Viet Nam and went from the Army to the Air Force side to fly fighter jets for the Missouri Air National Guard. He retired from the military with 23 years of service. Within the civilian arena of his aviation career, he flew nearly everything from turbo-props to jets. The last few years before retirement, Roger's normal route was from Houston, Texas to Narita, Japan and back to Houston.
He incorporated the flying protocol of using a checklist and planning to create his system. He also carried the skill of being a perfectionist as a pilot into the landscape-lighting business that created his checklist type system. Because of that system, he was able to write an awesome "how to" book on installing landscape-lighting systems.
He wanted to create a book that was simple and clear enough to share with anyone that is interested in installing there own system but did not know how to complete the task. His guarantee is anyone who buys, studies and follows his book, can install a system that will turn their home into a showcase masterpiece. To locate his book, just start with http://www.landscapelightingabc.com
Roger graduated from Southern Illinois University with a B.S. degree in Business Administration. After he left the farm in Illinois, he always was involved in business and started several companies throughout the years. Six years before retiring from the airlines, he started a landscape-lighting company that is doing extremely well today. The combination of business and flying caused Roger to create a system that anyone can follow and be successful installing landscape lighting for their home.
For more information, visit Neumann's site at http://www.abclandscapelighting.com
Article Source: ArticlesBase.com
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Use the constant voltage drop model to find..?
http://imageshack.us/photo/my-images/515/97784423.jpg/
Thats the problem im working with, were given Vs= 6v, R1= 220 ohm, R2= 220 ohm and 'the diode on voltage Von= 0.7V. ' Use the constant voltage drop model to find Id [answer in mA]
I found the thevenin equivalent for a- b which is 3v, but i dont understand how to use the voltage drop model to find iD.. any help would be great !
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What would the voltage drop from this parallel resistor be?
There is 3 parallel resistors with the total voltage being 150V and the total Resistance being 5 Ohms. What would the voltage be across one of the parallel resistors being 4k Ohms.
The information that i have been given is:
VT = 150V
IT = 30A
RT = 5 Ohms
I1 = 5A
R2 = 4k Ohms
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electric fan confusion?
a fan us running taking a constant power. if the voltage drops then what will happen to the current taken up by the fan.
A. will it increase
B. will it decrease
C. will it remain the same
answer this anyone with a reason. am not getting.
even am confused with both the answers. wil it increase or decrease.
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How to calculate the velocity of an electron accelerated through a voltage drop?
I'm can't figure out how to calculate the velocity of an electron after it has been accelerated by a voltage drop. Specifically, an electron from rest accelerated through a Gigavolt at which it travels at constant velocity.
Any help is appreciated, thanks
Yes, I am studying relativity. I missed a day, and looking at the notes it would seem that at this level physicists start referring to every quantity in forms of eV... which I thought was kind of weird.
So, I would use the relativistic energy equation, E = mc^2 + 1/2mv^2, where mc^2 is the rest energy
1 Gigavolt - rest energy = 1/2mv^2 and solve for v?
And, I guess might be unrelated to the initial equation, but can I assume the rest mass of a particle is the every-day one used in the lab frame?
Thanks,
Ah, I didn't see your addition there. Thank you.
I guess I had the wrong equation up there... when would I consider using that one?
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i want to fire the scr with push button switch.......?
i want to fire the scr whose anode voltage is at 24 vdc and cathode is connected to 24 vdc relay, now i think common value of igt in all scr are around 25 ma and gate voltage of 1.5 volt. now i want to make a voltage divider to get 1.5 v from the same 24 dc applied to scr's anode suppose two voltage divider resistors are r1 and r2 and want to get 1.5 volt acroos r2 second end of r2 is connected to ground now from the another end of r2 at which the voltage drop is 1.5 v i connect it to gate of scr, so what should be the values of resistor, whether i have to connect the resistor in series with gate to limit gate current????
i want to trigger the scr by push button swith placed between 1.5 v and gate
please explain the arrangement of circuit..
if any other better idea please suggest me, i don't want to damage scr
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I thought V-Drop is V * I...so why isnt this working?
http://postimage.org/image/je01vvfsh/
R1= 18, R2= 300, R3= 200, R4= 90
I have calculated the following
Current Through R1= .424 A
Voltage Drop Across R1= 7.632 V
Power Dissipation in R1= 3.23 W
Current Through R2= .077A
Current Through R3= .077 A
Current Through R4= .347
I need voltage across R2,R3, and R4. when i do V* I it doesnt work but it works for R1
I mean R * I.
Doing 300 * .077 does not work for the voltage drop across R2. why is that?
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Fan assisted oven wired in 1.5mm cable?
I carried out a cable sizing and volt drop calculation for a 2.8kw oven wired in twin and earth + cpc using
a BS 60898 fuse,
ambient temp 35 degrees
No thermal insulation
No grouping factor
I used Ib = P/V = 2800/230 = 12.17 (In)
Which means a 15A breaker can be used correct?
I then done
It = In/ CAxCGxCIxCC
This came to 15/0.94 as no other factors were used except Ca = 0.94 which gave me 15.95A and using the regs book it showed that a 1.5mm cable could be used to supply it? Even after calculating the voltage drop which was 5.99V which is within 5% of 230 it was still acceptable?
Im an apprentice electrician so I have some knowledge to this stuff but a 1.5mm cable run off a 15A breaker doesnt seem right?
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