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10.8V now 12V worldwide

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Biggie    118

Just like the rest of us, reviewers can't wait forever for dewalt to release new tools haha.

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kornomaniac    466
1 hour ago, Framer joe said:

Again, protool reviews said the same thing when the Makita rear came out. But real world tests all over you tube show the Flexvolt as more powerful and more cuts....the guy using the saw is the subjective part...of course pro tool reviews has two guys. Neither of which are contractors or framers,but instead two review hacks paid by manufacturers to review tools.........

.        ...  (.6) loss by the Flexvolt ,really? ..both were tie in all tests...again Makita hit a homerun with the ..rearhandle saw...great saw and I'll say it again ,,one of the most innovative companies, can't go wrong with them....

.       ...also love how these reviewers have test before Dewalt releases their new tools, happens every time..........I use Makita tools everyday at work framing along with Dewalt..................one note: Dewalt designs their tools to be used with their accessories. Not a random blade chosen...it Does make a huge difference in speed and cuts, just as when you use a certain bit on an sds hammer ,,,completely different result.....no other manufacturer does this...Dewalt is a system ,not just a tool,.     .....have a great day ! Love your Makita passion, ! 

 

Well too each his own. We're both passionate about our brands. Marketing goes a long way ;)

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Pouet    17
15 hours ago, kornomaniac said:

 

 As always  power and runtime are linked too eachother. Make a tool more powerfull and you'll limit runtime. Go for extended runtimes and you'll lack power's because batteries only have so much energy that needs to be distributed between either power or runtime.

 

It's not that clear cut. A more powerful tool will finish most of the time its job faster so you don't end up using more energy (of course some of it is loss through extra heat). The trade off is more between weight and power in my opinion.

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kornomaniac    466
2 minutes ago, Pouet said:

 

It's not that clear cut. A more powerful tool will finish most of the time its job faster so you don't end up using more energy (of course some of it is loss through extra heat). The trade off is more between weight and power in my opinion.

 

Power(watt) = voltage * amps.  That formula doesn't change because your tool has your favourite colour.

 

If you make a tools more powerfull ( to finish the job faster as you said) you either going to need more volts or more amps or both. 

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Stercorarius    2,496
1 hour ago, kornomaniac said:

 

Power(watt) = voltage * amps.  That formula doesn't change because your tool has your favourite colour.

 

If you make a tools more powerfull ( to finish the job faster as you said) you either going to need more volts or more amps or both. 

 

Or make them more efficient. You really like ohm's law like everything uses energy in equal efficiency, or that it's the only judge of how well a tool performs.

 

Yeah you can tow a 7 ton trailer across the United States of America with a Toyota Tacoma that has a 100 gallon tank and say driving there requires x joules of work.

 

One could say, "While no matter what you're favorite color of vehicle there's no changing that gasoline has 46.4 MJ/kg so that means that that Tacoma has the potential to do 13131.2 MJ of work.

 

Your F250 is diesel but that's all marketing wank and people aren't smart enough to see through it unless they're as smart as me. Doesn't matter how much of a fanboy you are, it doesn't change that diesel has 48 MJ/kg and and an F250 has a 35 gallon tank so that means you only have 5049.6 MJ of work available.

 

13131.2 MJ is a larger number than 5049.6 MJ. It's so simple even you should be able to see it. There's no getting around that no matter how big of a hard on you have for an F250 or how good Ford's marketing department is. A Tacoma is the better choice."

 

Good luck with that. Would you install low voltage single phase motors when high voltage three phase is also available? I know it's not as much of a difference with cordless tools.

 

Yes ohm's law accurately conveys potential energy of a battery but don't go whipping it around like that's the criteria for testing tools. If you truly believe that then I'll trade you a corded Ryobi reciprocating saw for your 2x18 cordless Makita, or even a brushed cordless grinder with a 9ah for your brushless grinder with a 5ah. Hey

Power (Watt)=voltage*amps and the first one has more of both and the second one has same voltage and more amps. Obviously they're better tools.

 

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kornomaniac    466

@Stercorarius

 

We're not talking about trucks and we're not talking about corded tools.

 

Keep the discussion and comparisons between cordless tools and then 4/5th of your post was unnecessary

 

It's a very simple statement: batteries have a limited amount of energy. And every cordless tool is a balance act between power and runtime. 

 

If you make your tool stronger/faster/more powerfull you'll use up more energy. 

 

That's all that statement said in the case of cordless tools like what we're discussing on the forum.

 

 

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Stercorarius    2,496
57 minutes ago, kornomaniac said:

@Stercorarius

 

We're not talking about trucks and we're not talking about corded tools.

 

Keep the discussion and comparisons between cordless tools and then 4/5th of your post was unnecessary

 

It's a very simple statement: batteries have a limited amount of energy. And every cordless tool is a balance act between power and runtime. 

 

If you make your tool stronger/faster/more powerfull you'll use up more energy. 

 

That's all that statement said in the case of cordless tools like what we're discussing on the forum.

 

 

Sorry I was just trying to demonstrate that you can't really use one principal to judge a whole tool let me edit that a little for you.

 

 

Or make them more efficient. You really like ohm's law like everything uses energy in equal efficiency, or that it's the only judge of how well a tool performs.

 

 

Yes ohm's law accurately conveys potential energy of a battery but don't go whipping it around like that's the criteria for testing tools. If you truly believe that then I'll trade you a brushed cordless grinder with a 9ah for your brushless grinder with a 5ah. Hey

Power (Watt)=voltage*amps and it has  the same voltage and more amps. Obviously it's a better tool, right?

 

 

If you make your tool stronger/faster/more powerfull you'll use up more energy. 

 

Really you want to stand by that? If you make a tool stronger/faster/more powerful you'll put out more energy. Does that often mean you use more, yes, but not always. I'm not saying that I have a design for a system that defies the law of conservation at all. I'm saying energy stored in any form can be put into use more efficiently and get more useable energy while draining less of the stored energy. Why would you get offered rebates for converting to led lighting? Sorry not about cordless tools don't want to confuse you. Let's say instead you have two cordless lights one draws x amps at 10.8v nominal and uses an incandescent bulb, the other draws the same x amps at the same 10.8v nominal and uses an LED light. Are you going to tell me those lights are both just as powerful? Let's talk impact drivers. You are then saying that because a brushed DeWalt impact driver draws more amps at 18v it is more powerful than your td170 because it draws less amps at 18v? You said more amps or voltage or both is the only way to make something, "stronger/faster/more powerfull" Your td170 doesn't draw more than brushed impact drivers at 18v then if your statement is correct you have an impact driver that is inferior to a brushed DeWalt impact. Even not all brushless motors are created equal. Why would Makita release so many impact drivers so frequently if they weren't improving the motors? Makita, DeWalt, Bosch, Milwaukee have until recently been using 18650 cells that discharge the same 18v at amperage rates that haven't had any huge increase. So has the power and runtime of tools also not increased? I'm fairly certain it has. You can very much have an increase in power and runtime while decreasing draw.

 

Don't worry I'm sure you would be a perfect fit for the Milwaukee marketing department if they're ever hiring.

 

 

 

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Framer joe    455

While I don't have the technical knowledge of @Stercorarius I can say I'm a Dewalt guy ,yes, but I use both brands at work. Also Milwaukee....In real world use ,the flex circ will outperform the rear handle  Makita or FUEL......I only care about actual daily abuse/use of tools...I see the Makita batteries draining much much faster then Dewalt on a comparable tool, ....I see my Dewalt cordless tools working hard and lasting years...

.     ...I'm there to make money,not baby a tool...sometimes these test people do show one tool .seconds better, that's not any advantage......an efficient,powerful  brushless motor in a durable frame is all I ask...I want the framing done as fast as possible......in my opinion the Flexvolt line of tools is ..Unmatched ...I think it's bad ass when a 60v tool at 3ah does more cuts then a 18v tool at 9ah....and much easier and faster......it's not Hype...cutting stair stringers  rafters and ripping lvl stock is extremely difficult for an 18v tool...they never show those cuts on you tube just a dry lumber,cross cut....

.      ...I will say the rear handle Makita is close, of course it won't be close at all to the rear handle 60v flex.................and the new Milwaukee 18v impact wrench is in a class it's own....1000ft pounds on,1400 ft pounds off...insane !!

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Pouet    17
6 hours ago, kornomaniac said:

 

Power(watt) = voltage * amps.  That formula doesn't change because your tool has your favourite colour.

 

If you make a tools more powerfull ( to finish the job faster as you said) you either going to need more volts or more amps or both. 

 

Listen I have a degree in electrical engineering. I think I know what is power is and I know what the scientific meaning of work is.

 

If your battery has 200 Wh of energy into it, you are going to produce 200 Wh of work out no matter what the voltage, if we are neglecting what you are loosing in term of heat. It doesn't matter what voltage your tool is. Your battery will always have the same amount of energy inside of it. You can't create new energy out of thin air just by switching the voltage.

 

If you are drilling the same hole with a 20V or a 60V drill, you are going to need the same amount of energy (again if we are neglecting heat losses which could be higher given a more powerful tool). The voltage has nothing to do with the energy required. One tool will complete the job faster than the other but your batteries will have lost the same amount of energy at the end of the job.  It's basic thermodynamics.  

 

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Pouet    17
3 hours ago, kornomaniac said:

@Stercorarius

 

We're not talking about trucks and we're not talking about corded tools.

 

Keep the discussion and comparisons between cordless tools and then 4/5th of your post was unnecessary

 

It's a very simple statement: batteries have a limited amount of energy. And every cordless tool is a balance act between power and runtime. 

 

If you make your tool stronger/faster/more powerfull you'll use up more energy. 

 

That's all that statement said in the case of cordless tools like what we're discussing on the forum.

 

 

 

You will use more energy but you will get the same amount of useful work out of the battery because Work = Power * runtime - heat losses. The same job will require the same amount of energy if we are neglecting heat losses. The only difference is one tool will finish faster than the other.

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dwain    790
9 hours ago, Framer joe said:

 pro tool reviews has two guys. Neither of which are contractors or framers,but instead two review hacks paid by manufacturers to review tools.........

 

Two things:

 

1) unless you have a good reason for saying so, that's a pretty unfair thing to say. As I've previously explained, tool reviewers receive tools from manufacturers for review. That doesn't make us biased. If you were getting free tools from one company and not the others ... that would be cause for concern. Even if a reviewer is charging money for reviews (I'm not sure if any do), if they charge the same to all companies, they may still be able to do so without bias. 

 

2) the reviewer/comparer of those saws is Michael Springer, and independant tool blogger who has been around a LONG time. I've met him personally at the recent Metabo tool factory in Germany. It was his 4th time there over the years! I can tell you that his tool knowledge is great, and he is a very practical guy who knows how to use tools.

 

You may have some small or large level of disdain for reviewers, but think harder. Full time contractors don't have time to do such comparisons/reviews. So, if reviewers didn't do them, they wouldn't exist.

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dwain    790

I think there is a little bit of nuance to the whole runtime v power discussion too.

 

- Total Wh matters a lot

- The number of cells and the output capacity (amps) of those cells is VERY important

- All cells will drain with less efficiency as they reach their upper-band temperatures

- some motors are inherently more efficient than others

- the size of the motor matters. a larger drill will have better efficiency for bigger tasks than a small drill. a small drill will have better efficiency for smaller holes.

- voltage is a factor. higher voltage means thinner cables, but thicker insulation (for instance). I'm not sure how much a factor voltage is, but it is A factor.

- the electronics and controls that limit the cells and tool play a large factor too

 

My point being that it is difficult to know just from specs. Real world test across a variety of applications (e.g. 2x4 pine cuts, 2x10 sleepers and then ripping LVLs) is the only real way to know.

 

Every manufacturer is going to choose the test that suits their tool best. 

 

 

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Pouet    17
14 minutes ago, dwain said:

I think there is a little bit of nuance to the whole runtime v power discussion too.

 

- Total Wh matters a lot

- The number of cells and the output capacity (amps) of those cells is VERY important

- All cells will drain with less efficiency as they reach their upper-band temperatures

- some motors are inherently more efficient than others

- the size of the motor matters. a larger drill will have better efficiency for bigger tasks than a small drill. a small drill will have better efficiency for smaller holes.

- voltage is a factor. higher voltage means thinner cables, but thicker insulation (for instance). I'm not sure how much a factor voltage is, but it is A factor.

- the electronics and controls that limit the cells and tool play a large factor too

 

My point being that it is difficult to know just from specs. Real world test across a variety of applications (e.g. 2x4 pine cuts, 2x10 sleepers and then ripping LVLs) is the only real way to know.

 

Every manufacturer is going to choose the test that suits their tool best. 

 

 

 

Exactly, the name of the game is efficiency which basically means reducing the amount of lost energy due to heat and there are many many factors involved in that equation. Picking a battery voltage is not about choosing between power or runtime but about balancing some variables like weight, power, ... given the expected use of the tool.

 

The reason why higher tension is potentially better is because energy losses in a circuit are equal to current² * resistance. Therefore, if you lower the current needed by rising the voltage, you are in theory supposed to get less energy lost in terms of heat. But again that's just one variable and there are other factors involved that need to be checked before coming to a conclusion.

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dwain    790
4 minutes ago, Pouet said:

Therefore, if you lower the current needed by rising the voltage, you are in theory supposed to get less energy lost in terms of heat. But again that's just one variable and there are other factors involved that need to be checked before coming to a conclusion.

 

But only if you maintain the cable size. We just don't have the kind of data (and possibly expertise also) necessary to have this debate scientifically.

 

Which is why I think real world tests are the way to go :)

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Pouet    17
16 minutes ago, dwain said:

 

But only if you maintain the cable size. We just don't have the kind of data (and possibly expertise also) necessary to have this debate scientifically.

 

Which is why I think real world tests are the way to go :)

 

The smaller cable size will rise the resistance so the equation is still valid. And I think it would be stupid to raise the voltage and then using smaller cables. Why give yourself all the trouble of using a new tension in the first place if you are going to use smaller cables?

 

But I totally agree with you that you can't judge a tool just by looking at the claimed numbers. That's why reviews are important. There are just too many factors at play. I have no opinion on which tool is better and nothing against Dewalt or Makita. But it's important to me that we don't spread false scientific statements whatever you may think of a brand.  

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jkeating3    63
1 hour ago, Pouet said:

 

Listen I have a degree in electrical engineering. I think I know what is power is and I know what the scientific meaning of work is.

 

If your battery has 200 Wh of energy into it, you are going to produce 200 Wh of work out no matter what the voltage, if we are neglecting what you are loosing in term of heat. It doesn't matter what voltage your tool is. Your battery will always have the same amount of energy inside of it. You can't create new energy out of thin air just by switching the voltage.

 

If you are drilling the same hole with a 20V or a 60V drill, you are going to need the same amount of energy (again if we are neglecting heat losses which could be higher given a more powerful tool). The voltage has nothing to do with the energy required. One tool will complete the job faster than the other but your batteries will have lost the same amount of energy at the end of the job.  It's basic thermodynamics.  

 

Considering your degree, I would assume your working knowledge is greater than mine, but ignoring losses due to heat seems like it wrecks the analysis.

 

Energy loss due to heat is energy that isn't doing work, and higher amperage is what causes that heat loss. So with higher voltage, you can use less amperage, which means less energy loss due to heat, which means more of that energy is going towards doing work. To me, it's not about a bigger tank, it's about not having a fuel leak.

 

All that is to say that higher voltage tools can push themselves harder without hemorrhaging power. But again, I defer to your expertise.

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Stercorarius    2,496
1 hour ago, Pouet said:

 

Listen I have a degree in electrical engineering. I think I know what is power is and I know what the scientific meaning of work is.

 

If your battery has 200 Wh of energy into it, you are going to produce 200 Wh of work out no matter what the voltage, if we are neglecting what you are loosing in term of heat. It doesn't matter what voltage your tool is. Your battery will always have the same amount of energy inside of it. You can't create new energy out of thin air just by switching the voltage.

 

If you are drilling the same hole with a 20V or a 60V drill, you are going to need the same amount of energy (again if we are neglecting heat losses which could be higher given a more powerful tool). The voltage has nothing to do with the energy required. One tool will complete the job faster than the other but your batteries will have lost the same amount of energy at the end of the job.  It's basic thermodynamics.  

 

Yes but higher voltage is more efficient. You have less resistance and less loss to heat is what I believe he was getting at. No one is saying that the initial potential energy in a battery is going to be changed by how it is drawn.

Edit:and that higher voltage while maintaining current does increase the amount of energy being delivered.

 

57 minutes ago, Pouet said:

 

Exactly, the name of the game is efficiency which basically means reducing the amount of lost energy due to heat and there are many many factors involved in that equation. Picking a battery voltage is not about choosing between power or runtime but about balancing some variables like weight, power, ... given the expected use of the tool.

 

The reason why higher tension is potentially better is because energy losses in a circuit are equal to current² * resistance. Therefore, if you lower the current needed by rising the voltage, you are in theory supposed to get less energy lost in terms of heat. But again that's just one variable and there are other factors involved that need to be checked before coming to a conclusion.

Just if he's confusing anyone by tension he means voltage. Why use the term tension instead? Couldn't tell you. 

 

35 minutes ago, Pouet said:

 

The smaller cable size will rise the resistance so the equation is still valid. And I think it would be stupid to raise the voltage and then using smaller cables. Why give yourself all the trouble of using a new voltage in the first place if you are going to use smaller cables?

So you can use smaller wire and other components. The components being where the gains are made. Wiring and components are a bigger deal when you're dealing with industrial wiring, but also applicable to the internals of power tools.

 

 

 

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Pouet    17
44 minutes ago, jkeating3 said:

Considering your degree, I would assume your working knowledge is greater than mine, but ignoring losses due to heat seems like it wrecks the analysis.

 

Energy loss due to heat is energy that isn't doing work, and higher amperage is what causes that heat loss. So with higher voltage, you can use less amperage, which means less energy loss due to heat, which means more of that energy is going towards doing work. To me, it's not about a bigger tank, it's about not having a fuel leak.

 

All that is to say that higher voltage tools can push themselves harder without hemorrhaging power. But again, I defer to your expertise.

 

I was neglecting heat losses to explain the concept that power,. runtime and work are linked together and not independent variables. I don't think any engineer would neglect heat losses. That would be crazy. That's what the job is all about. I wrote the same exact thing about higher voltages in another comment, that reducing current would mean fewer heat losses.

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Pouet    17
18 minutes ago, Stercorarius said:

Yes but higher voltage is more efficient. You have less resistance and less loss to heat is what I believe he was getting at. No one is saying that the initial potential energy in a battery is going to be changed by how it is drawn.

You don't have less resistance, you have less current. Take the time to read my comments again. I said higher voltage is in theory better because it reduces heat losses. I don't know why you are arguing with me about something I already said was true. What I said that's was wrong to say is that choosing a voltage is about choosing between runtime and power. It's not because runtime, power and work are not independent variables.

 

Quote

Just if he's confusing anyone by tension he means voltage. Why use the term tension instead? Couldn't tell you. 

 

Sorry if it's confusing anyone. Tension is the proper term and the one we were using in school but yes it's the same as voltage.

 

Quote

So you can use smaller wire and other components. The components being where the gains are made. Wiring and components are a bigger deal when you're dealing with industrial wiring, but also applicable to the internals of power tools.

 

 

 

 

Well ok then but if you use smaller wires, you increase the resistance and therefore the heat losses. I think Dewalt would prefer to increase the runtime then save a bit of money on smaller components. I could be wrong, I don't know what their costs are.

 

 

 

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Stercorarius    2,496
25 minutes ago, Pouet said:

You don't have less resistance, you have less current.  And less current means....... yeah less resistance.

 

 

Sorry if it's confusing anyone. Tension is the proper term and the one we were using in school but yes it's the same as voltage.

Not any more proper than voltage. Maybe in terms of engineering circuitry, I couldn't tell you there. In the context of any real world applications it's voltage. I'll let you have tension the next time I see any manufacturer or tradesman using it. It can be referred to as tension but was originally and still more commonly referred to as voltage. The only time you hear it referenced as tension is in high tension lines. I get that voltage produces tension on electrons so that's probably why you were told that way. Maybe it's just a regional thing to call it tension. If you can find one website that designates tension as being more proper I'll concede. 

 

 

Yes I am just an angry a-hole.

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comp56    5,655

super semi conductors .....just throwing that out there.....

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Framer joe    455

I love information.....very technical stuff .....what I do know For FACT ..IS..the Flexvolt Circ saw cuts more lumber of any dimension and rips more lumber then any other circ saw or rear handle saw on the market in any country....and the soon to be released Flexvolt rear handle will perform even better then that......I don't know why, that's not my world, but using these tools everyday 10hr a day is my world..well actually I just read plans, argue with architects.meet with engineers for redesign and run the crews...but I do use them every weekend building decks....

.      I think the 60v design so far is unmatched and can only get better, although the rear Makita is closer ,the FUEL circ isn't close.....and the new rear 60v will be ..as they say "game changer" ...

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dwain    790

^^ I also LOVE the FV saw. It is surprising to me that the 18x2 saw from Makita isn't comparable in runtime, but then again the original version isn't brushless is it?

 

I would have thought that a 9Ah FUEL Saw would compete on runtime for really light cutting, like melamine or 35mm studs. Can you confirm that it doesn't?

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dwain    790
3 hours ago, Pouet said:

 

The smaller cable size will rise the resistance so the equation is still valid. And I think it would be stupid to raise the voltage and then using smaller cables. Why give yourself all the trouble of using a new tension in the first place if you are going to use smaller cables?

 

But I totally agree with you that you can't judge a tool just by looking at the claimed numbers. That's why reviews are important. There are just too many factors at play. I have no opinion on which tool is better and nothing against Dewalt or Makita. But it's important to me that we don't spread false scientific statements whatever you may think of a brand.  

 

Isn't that EXACTLY why transmission mains are so high voltage, to keep cable size more practical?

 

I agree it's unlikely Dewalt have dropped the cable sizes from the standard size you see in most power tools, but its possible. Just saying we are debating about super technical engineering design without any data ....

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kornomaniac    466
5 hours ago, Pouet said:

 

You will use more energy but you will get the same amount of useful work out of the battery because Work = Power * runtime - heat losses. The same job will require the same amount of energy if we are neglecting heat losses. The only difference is one tool will finish faster than the other.

 @Pouet well that's what I've been saying isn't it?

 

Your first post said: ' a more powerfull motor will finish quicker and use less energy '.  Which I didn't agree on.

Which is Exactly the opposite from what you said later on :)

 

There is a case of a few percentage efficiency from going to higher voltages ( like flexvolt, less heat loss ) but all in all you repeatedly my initial statement: thesame job requires thesame amount of energy.

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