Whats The Wattage In Bkk?

[chownah]

quadricorrelator,

Bedlam is a forum here on TV but you have to have 500 posts before you are given access. This is to weed out the truly obnoxious and unruly posters who usually get eliminated before they reach 500. Mathematics is never discussed.

Back on topic.

At this link:http://www.miskatonic.org/godel.html

I found this which was taken from History of Mathematics, Boyer:

"Gödel showed that within a rigidly logical system such as Russell and Whitehead had developed for arithmetic, propositions can be formulated that are undecidable or undemonstrable within the axioms of the system. That is, within the system, there exist certain clear-cut statements that can neither be proved or disproved. ..........."

I have always thought that the concept of statement was more general than the concept of formula......formulas being a subset of the set of statements. In most discussion of this type I have seen people use the terms 'propositions' and 'statements' fairly interchangeable but I've never (that I can remember) heard anyone use the term 'formula' at this level of mathematical consideration.

[chownah]

Now could somebody please explain Ohm's Law.  

<{POST_SNAPBACK}>

Simple....if you hook up a battery with some wires to a resistor ( a resistor is something that resists the flow of electrons...duh?) and then you measure the voltage(the battery makes the voltage) and the resistance (the resistor makes the resistance) and the current (how many electrons are going buy in a second is how you measure current....using an ammeter perhaps) and you write these values down....like below

9 volts (a nine volt battery was used)

1 ohm (a one ohm resistor was used)

9 amps (thats a measure of how many electrons were passing...not 9 electrons silly....9 times some really really huge number which is equal to 1 amp)

then you change the battery and you find:

1.5 volts (a single d cell)

1 ohm (the same resistor)

1.5 amps (do you see some coincidence going on here?)

then you change the resistor:

1.5 volts (the same d cell)

0.5 ohm (smaller resistor here)

3.0 amps (does this make sense?)

And to make a long story not much longer....if you did this with alot of different batteries and resistors you would find the if you take the resistance (measured in ohms) and multiply it times the current (measured in amps) that the result would equal the voltage (measured in volts). Or E=IR (don't ask why but E stands for the voltage, I stands for the current and r is the easy one.) Simple, no? This is the explanation so that newbies will understand how Ohm's law is used....the explanation as to why this works is really interesting and truly elegant but to do it justice you really need to get a good physics book and read about it there....REALLY interesting stuff.

[chuchok]

Now could somebody please explain Ohm's Law.  

<{POST_SNAPBACK}>

Simple....if you hook up a battery with some wires to a resistor ( a resistor is something that resists the flow of electrons...duh?) and then you measure the voltage(the battery makes the voltage) and the resistance (the resistor makes the resistance) and the current (how many electrons are going buy in a second is how you measure current....using an ammeter perhaps) and you write these values down....like below

9 volts (a nine volt battery was used)

1 ohm (a one ohm resistor was used)

9 amps (thats a measure of how many electrons were passing...not 9 electrons silly....9 times some really really huge number which is equal to 1 amp)

then you change the battery and you find:

1.5 volts (a single d cell)

1 ohm (the same resistor)

1.5 amps (do you see some coincidence going on here?)

then you change the resistor:

1.5 volts (the same d cell)

0.5 ohm (smaller resistor here)

3.0 amps (does this make sense?)

And to make a long story not much longer....if you did this with alot of different batteries and resistors you would find the if you take the resistance (measured in ohms) and multiply it times the current (measured in amps) that the result would equal the voltage (measured in volts). Or E=IR (don't ask why but E stands for the voltage, I stands for the current and r is the easy one.) Simple, no? This is the explanation so that newbies will understand how Ohm's law is used....the explanation as to why this works is really interesting and truly elegant but to do it justice you really need to get a good physics book and read about it there....REALLY interesting stuff.

<{POST_SNAPBACK}>

So Chow ,if i'm using a 15 AMP skill saw that puts out max power of 2200watts, what would be the average voltage drop if your power cord was about 80' long?

[raro]

Back on topic. 

At this link:http://www.miskatonic.org/godel.html

<{POST_SNAPBACK}>

heyheyhey! The original topic was whether you should use 110 or 220 Amp fuses in bkk or something like that....where can I get Miller Light?

Edited November 6, 2005 by raro

[quadricorrelator]

quadricorrelator,

Bedlam is a forum here on TV but you have to have 500 posts before you are given access.  This is to weed out the truly obnoxious and unruly posters who usually get eliminated before they reach 500.  Mathematics is never discussed.

Back on topic. 

At this link:http://www.miskatonic.org/godel.html

I found this which was taken from History of Mathematics, Boyer:

"Gödel showed that within a rigidly logical system such as Russell and Whitehead had developed for arithmetic, propositions can be formulated that are undecidable or undemonstrable within the axioms of the system. That is, within the system, there exist certain clear-cut statements that can neither be proved or disproved. ..........."

I have always thought that the concept of statement was more general than the concept of formula......formulas being a subset of the set of statements.  In most discussion of this type I have seen people use the terms 'propositions' and 'statements' fairly interchangeable but I've never (that I can remember) heard anyone use the term 'formula' at this level of mathematical consideration.

<{POST_SNAPBACK}>

Thank you for the link. It looks interesting.

statement, sentence, proposition, formula: Statements and propositions and sentences all mean the same thing. Statements/propositions/sentences can be built up from other statements using the logical connectives: implies, and, or, not. The propositional or statement variables can only take on the value of true or false. The statement or proposition can be evaluated using truth tables for the connectives, when each variable is given a true or false assignment.

For example: "p implies q" is a proposition or statement. It's truth or falsity is as follows:

p q p implies q

t t t

t f f

f t t

f t t

But, you could have more complicated statements such as:

(p implies (q implies r)) implies ((p implies q) implies (p implies r))

(this statement is true for any of the eight true/false assignments given to its variables)

You can derive all statements from only one propositional axiom if you pick the right axiom.

The theory of deducing new statements/propositions from the axioms is called "The Propositional Calculus". The theory of determing the truth of statements/propositions/sentences is called "Propositional Logic"

The axioms of propositional calculus are complete. All deducible statements/propositions/sentences are true according to truth tables. And all statements/propositions/sentences which are true according to truth tables are deducible.

Formulas: A formula includes propostitions/statements/sentences, but also includes strings made up of the following additional objects: "for all", "there exists", predicates, functions, non-propositional variables, and constants.

Here is an example of a formula that is not a statement:

(there exists a y) (x < y)

In this case, x and y are non-propositional variables, "<" is a predicate.

You can turn this formula into a statement be preceding it with "for all x" or "there exists x"

You can convert any formula into a statement by adding enough "for all" or "there exists" in front of it. When you do, the statement can be evaluated to be true or false. When you do this, it is called the closure of a formula.

Here is another example (for all x) (x + y = 2)

x and y are variables, 2 is a constant, "+" is a function, "=" is a predicate

If you add the proper formulas to the axioms of the propositional calculus, then you get the full set of axioms for the Predicate Calculus. The Predicate Calculus is complete, but it is much more powerful and general than the Propositional Calculus. It is also complete.

The Predicate Calculus is the basis of Logic. You can create new theories by adding statements (but not formulas) to the axioms of the Predicate Calculus. For example, you can create Set Theory, Number Theory, Group Theory, etc. by adding the proper statements. The theory can be inconsistent or incomplete depending on which statements you use as axioms.

Godel proved one of his incompleteness theorems within Number Theory.

Set Theory contains the axioms for all of mathematics. Number Theory is contained within Set theory (as are all of the mathematical theories).

May I mention a website that I enjoy? It is the Metamathematics Proof Explorer Homepage. I can't find the address right now, but I will send it later.

[quadricorrelator]

quadricorrelator,

Bedlam is a forum here on TV but you have to have 500 posts before you are given access.  This is to weed out the truly obnoxious and unruly posters who usually get eliminated before they reach 500.  Mathematics is never discussed.

Back on topic. 

At this link:http://www.miskatonic.org/godel.html

I found this which was taken from History of Mathematics, Boyer:

"Gödel showed that within a rigidly logical system such as Russell and Whitehead had developed for arithmetic, propositions can be formulated that are undecidable or undemonstrable within the axioms of the system. That is, within the system, there exist certain clear-cut statements that can neither be proved or disproved. ..........."

I have always thought that the concept of statement was more general than the concept of formula......formulas being a subset of the set of statements.  In most discussion of this type I have seen people use the terms 'propositions' and 'statements' fairly interchangeable but I've never (that I can remember) heard anyone use the term 'formula' at this level of mathematical consideration.

<{POST_SNAPBACK}>

May I mention some sources I use to learn this material?

First I studied, "Introduction to Elementary Logic"

Now I am studying, "First Order Mathematical Logic" by Margaris and "Mathematical Logic" by Kleene.

All three books are Dover reprints so they are relatively cheap for texts. You can buy all three for about 1500 baht.

I found these books at Kinokuniya at the Emporium.

May I also suggest a website which I find very interesting: Metamathematics Proof Explorer Homepage. They start with the basic axioms of set theory and logic, and the derive some of the theorems of mathematics. The proof that 2 + 2 = 4 takes over 20,000 steps. It has over 2000 subtheorems. I think the deepest path back to the axioms is 129 levels.

[quadricorrelator]

Now could somebody please explain Ohm's Law.  

<{POST_SNAPBACK}>

Simple....if you hook up a battery with some wires to a resistor ( a resistor is something that resists the flow of electrons...duh?) and then you measure the voltage(the battery makes the voltage) and the resistance (the resistor makes the resistance) and the current (how many electrons are going buy in a second is how you measure current....using an ammeter perhaps) and you write these values down....like below

9 volts (a nine volt battery was used)

1 ohm (a one ohm resistor was used)

9 amps (thats a measure of how many electrons were passing...not 9 electrons silly....9 times some really really huge number which is equal to 1 amp)

then you change the battery and you find:

1.5 volts (a single d cell)

1 ohm (the same resistor)

1.5 amps (do you see some coincidence going on here?)

then you change the resistor:

1.5 volts (the same d cell)

0.5 ohm (smaller resistor here)

3.0 amps (does this make sense?)

And to make a long story not much longer....if you did this with alot of different batteries and resistors you would find the if you take the resistance (measured in ohms) and multiply it times the current (measured in amps) that the result would equal the voltage (measured in volts). Or E=IR (don't ask why but E stands for the voltage, I stands for the current and r is the easy one.) Simple, no? This is the explanation so that newbies will understand how Ohm's law is used....the explanation as to why this works is really interesting and truly elegant but to do it justice you really need to get a good physics book and read about it there....REALLY interesting stuff.

<{POST_SNAPBACK}>

So Chow ,if i'm using a 15 AMP skill saw that puts out max power of 2200watts, what would be the average voltage drop if your power cord was about 80' long?

<{POST_SNAPBACK}>

May I ask a question before going further? You say that the saw puts out a max power of 2200 watts. Do you mean that it consumes 2200 watts of electrical power, or that it puts out 2200 watts of mechanical power?

If it is consuming 2200 watts of electrical power (the first assumption above), then you can make calculations to determine the voltage drop in the power cord.

But, if you only know the mechanical output power (the second assumption above), then there is not enough information to answer your question. In this case, you would also have to know the electrical to mechanical conversion efficiency before you could answer the question.

Let me assume that the 2200 watts is the electrical power saw consumption (just for the purpose of going through the calculation).

In this case, you calculate the voltage across the saw is

v = p/i = 2200 watts /15 amps = 147 volts.

But, the voltage at the power outlet is 220 volts. So the drop in the line is 220 volts - 147 volts = 73 volts. This would mean that the power dissipated in the power cord was 73 volts * 15 amps = 1095 watts. The resistance of the cord would be

73 volts/ 15 amps = 4.87 ohms.

These results are unrealistic. I do not believe that an 80 foot power saw cord could have nearly 5 ohms of resistance. I do not believe that the power cord could dissipate 1095 watts. This would be the one of the worst power cord designs in history. The original assumption, that the saw is consuming 2200 watts of electrical power, must be incorrect.

However, if you can find the saw power consumption requirements (maybe it is on the saw somewhere), then we might be able to calculate more realistic results. I imagine that the power saw should come with power consumption requirements, or maybe you can find those requirements on the power saw company website.

I would expect that the drop in the power cord to be a small percentage of the total power used.

Edited November 6, 2005 by quadricorrelator

[ironwolf]

Man, I might Voltage. Dam this is way off subject.

[Crossy]

Man, I might Voltage. Dam this is way off subject.

Fun though isn't it?

And you have the answer to your original question, 220V 50Hz.

How's the 'passport' thread going?

[Jai Dee]

Man, I might Voltage. Dam this is way off subject.

Classic!!!

This has got to be one of the funniest threads I've read in a long time.

[chownah]

So Chow ,if i'm using a 15 AMP skill saw that puts out max power of 2200watts, what would be the average voltage drop if your power cord was about 80' long?

<{POST_SNAPBACK}>

First let me analyse the date.

15amp skill saw....my skill saw is 220 volt 6.1 amps so I'm going to assume that your saw is from the US and is running on 110 volts. If it was running on 220 volts then it would be monstrously huge...probably not realistic.

15amp x 110 volt = 1650 watts.....ooops, my assumption of 110 volts can not be correct because this would mean that it consumes less power then it outputs...this can not be. I'll change my assumption for the voltage to 220 volts.

15amp x 220 volts = 3300 watts = 4.4 hp (peak) That's a big skill saw.

I'll assume that the 15amp and the 2200 watt ratings are both taken at rated voltage (220volts) and at peak load.

80 foot long power cord....well....you could be using speaker wire for a power cord or you could be using 4 guage theatrical power cord....it would make a big difference which you used. I'll assume you are using a 18 guage cord since this is what I might use if I was going 80 feet...and 18 guage (copper) wire has 0.64 ohms per 100 feet so since the total circuit length in the cord is 160' then the resistance in the wire is 160 x 0.0064 = 1 ohm...I like easy numbers like '1'!!!

Now the hard part. First I want to caution that ohms law in the form E=Ir is really only meant for DC stuff and since this is AC the answer we get will be approximate but it will be a practical estimate never the less....I think.

Assume : saw is drawing 15 amp

then from E=Ir......220volts = 15 amps x saw's resistence....if you solve this you will find that the saw has about 14.7 ohms of resistence when at full load and at 220 volts.

add the resistence from the cord to the resistence of the saw...14.7 + 1.0 = 15.7....this is our first estimate of the resistence of the entire circuit with the saw running full tilt.

Then from E = Ir ...we get 220v = I x 15.7 ohms.....if you solve this you get that the first approximation of the current flowing in the entire circuit with the saw is only 14.0 amps....and since the saw has approximately 14.7 ohms and the wire has 1 ohm then since the voltage drop is directly proportional to the fraction of the resistence in the circuit element, the saw actually has a voltage drop of 220 volts x (14.7/15.7) = 206 volts.

So, our answer is APPROXIMATELY 206 volts of voltage drop across the saw when attached to an 80' extension cord made from 18 guage copper wire when the supply voltage is at 220 volts. I want to mention that this is not actually the correct answer....this answer assumes that the resistence of the saw is constant over its entire power range which is definitely not true. Also notice that the saw will only be drawing about 14 amps so it will not be producing its peak power as would be expected since the extension cord will reduce the voltage available to the saw below its rated value.

[SiamOne]

Is the wattage in BKK is it 220 or 110?

<{POST_SNAPBACK}>

don't worry IronWolf, people love to laugh at mistakes!!

Mind you, the wattage question is interesting as the electric companies let you choose what (?) you want, and charge accordingly

so if you want to fit a new aircon or waterheater, you may need to uprate your wattage. That probably means new, heavier wires to compensate

why is the voltage not 240 anyway, anyone??

[Crossy]

why is the voltage not 240 anyway, anyone??

Because it's 220V, who knows why they chose 220 as opposed to 230 or 210 (very old UK system) or even 110V or 100V (parts of Japan).

AFAIK only the UK was using 240V (now 230V to comply with European harmonisation).

UK kit will work nicely on 220 and 240V lamps will last longer.

[chuchok]

So Chow ,if i'm using a 15 AMP skill saw that puts out max power of 2200watts, what would be the average voltage drop if your power cord was about 80' long?

<{POST_SNAPBACK}>

First let me analyse the date.

15amp skill saw....my skill saw is 220 volt 6.1 amps so I'm going to assume that your saw is from the US and is running on 110 volts. If it was running on 220 volts then it would be monstrously huge...probably not realistic.

15amp x 110 volt = 1650 watts.....ooops, my assumption of 110 volts can not be correct because this would mean that it consumes less power then it outputs...this can not be. I'll change my assumption for the voltage to 220 volts.

15amp x 220 volts = 3300 watts = 4.4 hp (peak) That's a big skill saw.

I'll assume that the 15amp and the 2200 watt ratings are both taken at rated voltage (220volts) and at peak load.

80 foot long power cord....well....you could be using speaker wire for a power cord or you could be using 4 guage theatrical power cord....it would make a big difference which you used. I'll assume you are using a 18 guage cord since this is what I might use if I was going 80 feet...and 18 guage (copper) wire has 0.64 ohms per 100 feet so since the total circuit length in the cord is 160' then the resistance in the wire is 160 x 0.0064 = 1 ohm...I like easy numbers like '1'!!!

Now the hard part. First I want to caution that ohms law in the form E=Ir is really only meant for DC stuff and since this is AC the answer we get will be approximate but it will be a practical estimate never the less....I think.

Assume : saw is drawing 15 amp

then from E=Ir......220volts = 15 amps x saw's resistence....if you solve this you will find that the saw has about 14.7 ohms of resistence when at full load and at 220 volts.

add the resistence from the cord to the resistence of the saw...14.7 + 1.0 = 15.7....this is our first estimate of the resistence of the entire circuit with the saw running full tilt.

Then from E = Ir ...we get 220v = I x 15.7 ohms.....if you solve this you get that the first approximation of the current flowing in the entire circuit with the saw is only 14.0 amps....and since the saw has approximately 14.7 ohms and the wire has 1 ohm then since the voltage drop is directly proportional to the fraction of the resistence in the circuit element, the saw actually has a voltage drop of 220 volts x (14.7/15.7) = 206 volts.

So, our answer is APPROXIMATELY 206 volts of voltage drop across the saw when attached to an 80' extension cord made from 18 guage copper wire when the supply voltage is at 220 volts. I want to mention that this is not actually the correct answer....this answer assumes that the resistence of the saw is constant over its entire power range which is definitely not true. Also notice that the saw will only be drawing about 14 amps so it will not be producing its peak power as would be expected since the extension cord will reduce the voltage available to the saw below its rated value.

<{POST_SNAPBACK}>

What if it's a really, really hot day?

[raro]

Is the wattage in BKK is it 220 or 110?

<{POST_SNAPBACK}>

don't worry IronWolf, people love to laugh at mistakes!!

Mind you, the wattage question is interesting as the electric companies let you choose what (?) you want, and charge accordingly

so if you want to fit a new aircon or waterheater, you may need to uprate your wattage. That probably means new, heavier wires to compensate

why is the voltage not 240 anyway, anyone??

<{POST_SNAPBACK}>

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

Chownah, Quadricorrelator: May I ask what you guys do for a living?

Cheers

raro

[Tywais]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

Chownah, Quadricorrelator: May I ask what you guys do for a living?

Cheers

raro

BTW Integral of sine is 0 (zero). The peak of 220V RMS is 311V

[raro]

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

Chownah, Quadricorrelator: May I ask what you guys do for a living?

Cheers

raro

<{POST_SNAPBACK}>

BTW Integral of sine is 0 (zero). The peak of 220V RMS is 311V

<{POST_SNAPBACK}>

that's right, but since you have a negative voltage as well, you have to integrate over the modulus.

[chownah]

Is the wattage in BKK is it 220 or 110?

<{POST_SNAPBACK}>

don't worry IronWolf, people love to laugh at mistakes!!

Mind you, the wattage question is interesting as the electric companies let you choose what (?) you want, and charge accordingly

so if you want to fit a new aircon or waterheater, you may need to uprate your wattage. That probably means new, heavier wires to compensate

why is the voltage not 240 anyway, anyone??

<{POST_SNAPBACK}>

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

Chownah, Quadricorrelator: May I ask what you guys do for a living?

Cheers

raro

<{POST_SNAPBACK}>

I'm a rice farmer in the north.

[chownah]

So Chow ,if i'm using a 15 AMP skill saw that puts out max power of 2200watts, what would be the average voltage drop if your power cord was about 80' long?

<{POST_SNAPBACK}>

First let me analyse the date.

15amp skill saw....my skill saw is 220 volt 6.1 amps so I'm going to assume that your saw is from the US and is running on 110 volts. If it was running on 220 volts then it would be monstrously huge...probably not realistic.

15amp x 110 volt = 1650 watts.....ooops, my assumption of 110 volts can not be correct because this would mean that it consumes less power then it outputs...this can not be. I'll change my assumption for the voltage to 220 volts.

15amp x 220 volts = 3300 watts = 4.4 hp (peak) That's a big skill saw.

I'll assume that the 15amp and the 2200 watt ratings are both taken at rated voltage (220volts) and at peak load.

80 foot long power cord....well....you could be using speaker wire for a power cord or you could be using 4 guage theatrical power cord....it would make a big difference which you used. I'll assume you are using a 18 guage cord since this is what I might use if I was going 80 feet...and 18 guage (copper) wire has 0.64 ohms per 100 feet so since the total circuit length in the cord is 160' then the resistance in the wire is 160 x 0.0064 = 1 ohm...I like easy numbers like '1'!!!

Now the hard part. First I want to caution that ohms law in the form E=Ir is really only meant for DC stuff and since this is AC the answer we get will be approximate but it will be a practical estimate never the less....I think.

Assume : saw is drawing 15 amp

then from E=Ir......220volts = 15 amps x saw's resistence....if you solve this you will find that the saw has about 14.7 ohms of resistence when at full load and at 220 volts.

add the resistence from the cord to the resistence of the saw...14.7 + 1.0 = 15.7....this is our first estimate of the resistence of the entire circuit with the saw running full tilt.

Then from E = Ir ...we get 220v = I x 15.7 ohms.....if you solve this you get that the first approximation of the current flowing in the entire circuit with the saw is only 14.0 amps....and since the saw has approximately 14.7 ohms and the wire has 1 ohm then since the voltage drop is directly proportional to the fraction of the resistence in the circuit element, the saw actually has a voltage drop of 220 volts x (14.7/15.7) = 206 volts.

So, our answer is APPROXIMATELY 206 volts of voltage drop across the saw when attached to an 80' extension cord made from 18 guage copper wire when the supply voltage is at 220 volts. I want to mention that this is not actually the correct answer....this answer assumes that the resistence of the saw is constant over its entire power range which is definitely not true. Also notice that the saw will only be drawing about 14 amps so it will not be producing its peak power as would be expected since the extension cord will reduce the voltage available to the saw below its rated value.

<{POST_SNAPBACK}>

What if it's a really, really hot day?

<{POST_SNAPBACK}>

If its a really really hot day then my skill saw usually doesn't work at all!!!

[quadricorrelator]

Is the wattage in BKK is it 220 or 110?

<{POST_SNAPBACK}>

don't worry IronWolf, people love to laugh at mistakes!!

Mind you, the wattage question is interesting as the electric companies let you choose what (?) you want, and charge accordingly

so if you want to fit a new aircon or waterheater, you may need to uprate your wattage. That probably means new, heavier wires to compensate

why is the voltage not 240 anyway, anyone??

<{POST_SNAPBACK}>

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

Chownah, Quadricorrelator: May I ask what you guys do for a living?

Cheers

raro

<{POST_SNAPBACK}>

I am retired. I used to be an analog microchip designer.

[WhiteShiva]

Jeeeeezezzzz - Can we get back on the topic, people - please!!!!

He didn't ask about volts and sin curves. He asked about wattage!

Help the poor guy out - tell him the wattage in Bangkok. I'd say it is around

3.14 * sin(x^2) * log (t/Sqrt(1-v^2/c^2)) + bah ^ beer

where

x = no of electrical outlets in your house (or hotel room, if a tourist)

t = time spent in Thailand

v = velocity of your taxi when stuck in a Bangkok rot tit

c = speed of light (daytime!)

bah = stress factor

beer = beer

In other word, pretty d@mn close to 42.

What was the question, again?

[Tywais]

Jeeeeezezzzz - Can we get back on the topic, people - please!!!! 

He didn't ask about volts and sin curves.  He asked about wattage! 

Help the poor guy out - tell him the wattage in Bangkok. I'd say it is around

3.14 * sin(x^2) * log (t/Sqrt(1-v^2/c^2)) + bah ^ beer

where

x = no of electrical outlets in your house (or hotel room, if a tourist)

t = time spent in Thailand

v = velocity of your taxi when stuck in a Bangkok rot tit

c = speed of light (daytime!)

bah = stress factor

beer = beer

In other word, pretty d@mn close to 42.

What was the question, again?

One parameter you left out of the equation is "BG" (bar girl) and a lot of watts could be expended on this variable alone.

[dr_Pat_Pong]

Geez, there's a lot of bright sparks around here... 

<{POST_SNAPBACK}>

Shocking isn't it ?

[chuchok]

So Chow ,if i'm using a 15 AMP skill saw that puts out max power of 2200watts, what would be the average voltage drop if your power cord was about 80' long?

<{POST_SNAPBACK}>

First let me analyse the date.

15amp skill saw....my skill saw is 220 volt 6.1 amps so I'm going to assume that your saw is from the US and is running on 110 volts. If it was running on 220 volts then it would be monstrously huge...probably not realistic.

15amp x 110 volt = 1650 watts.....ooops, my assumption of 110 volts can not be correct because this would mean that it consumes less power then it outputs...this can not be. I'll change my assumption for the voltage to 220 volts.

15amp x 220 volts = 3300 watts = 4.4 hp (peak) That's a big skill saw.

I'll assume that the 15amp and the 2200 watt ratings are both taken at rated voltage (220volts) and at peak load.

80 foot long power cord....well....you could be using speaker wire for a power cord or you could be using 4 guage theatrical power cord....it would make a big difference which you used. I'll assume you are using a 18 guage cord since this is what I might use if I was going 80 feet...and 18 guage (copper) wire has 0.64 ohms per 100 feet so since the total circuit length in the cord is 160' then the resistance in the wire is 160 x 0.0064 = 1 ohm...I like easy numbers like '1'!!!

Now the hard part. First I want to caution that ohms law in the form E=Ir is really only meant for DC stuff and since this is AC the answer we get will be approximate but it will be a practical estimate never the less....I think.

Assume : saw is drawing 15 amp

then from E=Ir......220volts = 15 amps x saw's resistence....if you solve this you will find that the saw has about 14.7 ohms of resistence when at full load and at 220 volts.

add the resistence from the cord to the resistence of the saw...14.7 + 1.0 = 15.7....this is our first estimate of the resistence of the entire circuit with the saw running full tilt.

Then from E = Ir ...we get 220v = I x 15.7 ohms.....if you solve this you get that the first approximation of the current flowing in the entire circuit with the saw is only 14.0 amps....and since the saw has approximately 14.7 ohms and the wire has 1 ohm then since the voltage drop is directly proportional to the fraction of the resistence in the circuit element, the saw actually has a voltage drop of 220 volts x (14.7/15.7) = 206 volts.

So, our answer is APPROXIMATELY 206 volts of voltage drop across the saw when attached to an 80' extension cord made from 18 guage copper wire when the supply voltage is at 220 volts. I want to mention that this is not actually the correct answer....this answer assumes that the resistence of the saw is constant over its entire power range which is definitely not true. Also notice that the saw will only be drawing about 14 amps so it will not be producing its peak power as would be expected since the extension cord will reduce the voltage available to the saw below its rated value.

<{POST_SNAPBACK}>

What if it's a really, really hot day?

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If its a really really hot day then my skill saw usually doesn't work at all!!!

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Is that because heat will increase resistance, or that you are out in the fields?

[Thetyim]

Nope, you have all got it wrong, completely wrong.

He wants to know about wattage in Bangkok or

How many temples are there in Bangkok

At a guess, about 400

[mijan24]

Is the wattage in BKK is it 220 or 110?

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don't worry IronWolf, people love to laugh at mistakes!!

Mind you, the wattage question is interesting as the electric companies let you choose what (?) you want, and charge accordingly

so if you want to fit a new aircon or waterheater, you may need to uprate your wattage. That probably means new, heavier wires to compensate

why is the voltage not 240 anyway, anyone??

<{POST_SNAPBACK}>

Since we engulf in peacounting: As we talk Alternating Currents, the voltage is at any given time anywhere between zero and 360, depending on which point of the sinus curve you measure. The integral over a full period, however, is 220 to 240 volts.

Chownah, Quadricorrelator: May I ask what you guys do for a living?

Cheers

raro

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I am retired. I used to be an analog microchip designer.

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quadricorrelator as a analog microchip designer I am sure with all that air con you would have had a lot of sinusitis but when you installed the A/C and extra water heaters they would have preferred if you had increased the size of the wires carrying the current (as stated) and installed larger fuses 35amp to accommodate the extra elec flowing thru them.

Elec can be equated to water, pressure = volts, flow = current, pipe diamater = resistance - I used to be a garboligist then on promotion moved as supervisor saniation regurgitation - culminating an illusterous career with the NCC as head of the janitorial department with approx 25 brooms @ my disposal

Watt have I said all my secrets out of the disposal unit)

Edited November 8, 2005 by mijan24

[SiamOne]

I am retired.  I used to be an analog microchip designer.

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Lucky you retired before all that digital stuff took over!!

[mijan24]

I am retired.  I used to be an analog microchip designer.

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Lucky you retired before all that digital stuff took over!!

<{POST_SNAPBACK}>

That then begs the question Watt are you doing in Thailand?

Answer: Kicking off the digital redesine Hub!

[quadricorrelator]

I am retired.  I used to be an analog microchip designer.

<{POST_SNAPBACK}>

Lucky you retired before all that digital stuff took over!!

<{POST_SNAPBACK}>

It seems like analog should be dead, but somehow it is still alive. Many thought analog was on its way out 26 years ago (when I first started). But, the need for analog designers always seems to come up: power supply design for notebook computers is a hot field, cell phone radio frequency amplifiers is a hot field, disk drive design has quite a bit of analog circuitry in it, front end amplifier circuitry for fiber optics requires analog, phase lock loop design for communications, and on and on.

Data received from the world is analog. Such systems always need analog signal processing.

Maxim Corporation, Linear Technology Corporation, and the analog division of National Semiconductor are all analog companies which have been very profitable over a long period of time. But, they have recruiting difficulties.

There is a shortage of analog designers relative to the demand. Universities don't want to teach it because the world needs much greater numbers of digital designers and computer programmers. Students don't want to study it because it is not glamorous (at least not like digital design), it has a lot of mathematics in it, it takes about three or four years of industrial experience before you are independent enough to do a design on your own (partially because it is a mature field it takes time to learn everything already known), and it doesn't fit most people's temperment because it requires a very narrow focus (you spend your career learning more and more about the same basic devices; transistor, capacitor, resistor, and inductor).

But, when I was making my career decision, I didn't even consider digital design. It just wasn't for me. I like the math of analog circuitry. I liked the narrow focus. I liked the idea of going deeper and deeper into the same field.

I think there is quite a bit of job security for analog designers. I don't know how the pay scales compare for equivalent levels of competence.

Edited November 8, 2005 by quadricorrelator

[mijan24]

I am retired.  I used to be an analog microchip designer.

<{POST_SNAPBACK}>

Lucky you retired before all that digital stuff took over!!

<{POST_SNAPBACK}>

It seems like analog should be dead, but somehow it is still alive. Many thought analog was on its way out 26 years ago (when I first started). But, the need for analog designers always seems to come up: power supply design for notebook computers is a hot field, cell phone radio frequency amplifiers is a hot field, disk drive design has quite a bit of analog circuitry in it, front end amplifier circuitry for fiber optics requires analog, phase lock loop design for communications, and on and on.

Data received from the world is analog. Such systems always need analog signal processing.

Maxim Corporation, Linear Technology Corporation, and the analog division of National Semiconductor are all analog companies which have been very profitable over a long period of time. But, they have recruiting difficulties.

There is a shortage of analog designers relative to the demand. Universities don't want to teach it because the world needs much greater numbers of digital designers and computer programmers. Students don't want to study it because it is not glamorous (at least not like digital design), it has a lot of mathematics in it, it takes about three or four years of industrial experience before you are independent enough to do a design on your own (partially because it is a mature field it takes time to learn everything already known), and it doesn't fit most people's temperment because it requires a very narrow focus (you spend your career learning more and more about the same basic devices; transistor, capacitor, resistor, and inductor).

But, when I was making my career decision, I didn't even consider digital design. It just wasn't for me. I like the math of analog circuitry. I liked the narrow focus. I liked the idea of going deeper and deeper into the same field.

I think there is quite a bit of job security for analog designers. I don't know how the pay scales compare for equivalent levels of competence.

<{POST_SNAPBACK}>

quadricorrelator Good Post - thank you, I for one found it interesting, proving your never to old to learn something new. One of the most interesting points for me was the "narrow focus" there are many people who try to breakaway from a natural or taught narrow focus not knowing (I include myself) there are fields where it can be advantageous.

mijan24

Edited November 8, 2005 by mijan24