Just to add to the confusion, I have an electric lawn edger advertised as being a certain horsepower.
Reading the fine print, the (horse)power was calculated by taking the stall torque and multiplying by the free-running speed.
That’s enough to make any electrical engineer start foaming at the mouth.
Obvious bullshit like that notwithstanding, I think the trend is to report performance using more or less comprehensible, usually metric, figures. For instance, according to this page a hybrid BMW i8 has a turbocharged 3-cylinder engine providing 170 kW + an electric engine providing 96 kW, battery capacity 11.6 kW·h
Well Carribean_cruiser, this is what you get when you ask about certain things. Competing metaphors, and piddly distinctions that cloud your brain.
The closest I ever got to an explanation for this stuff that made sense was an old cartoon starring Augie Ampere and Victor Volt. I know that electricity comes into your house through a wire, but how it behaves when it gets there is beyond my ken. So they wire your kitchen with multiple circuits so you don’t overload any of them - dude, it’s the same electricity, it doesn’t care how you route it!
It’s too late for me, and I suspect you. Well-meaning people cannot hope to dissolve the barriers to our understanding, and I’ve long given up trying to get a grip on why, for instance, you need three prongs on some plugs and only two on others, and those two will have a wide prong and a narrow prong so you don’t plug it in the wrong way even though AC goes back-and-forth so why would it matter?
Do not attempt to clarify this, I won’t read it. I used to care but not any more. My big brother gets this stuff but he knows I never will.
Each ‘circuit’ they routed has its own breaker, typically a 15A breaker. If you had three kitchen circuits, you certainly could have run one circuit instead, with a 45A breaker. But then you’d have to upgrade all the electricsl wiring to the kitchen to handle 45A instead of 15. And, if any device misbehaves and trips the breaker, you would lose all the power to the kitchen instead of just the one circuut. So it’s safer, cheaper, and more fault-tolerant to divide the home into a number of lower-current circuits, each with their own breaker.
Fun fact - when McD’s came out with the quarter-pounder, a competing chain came out with a third-pounder for the same price - but it didn’t sell. Apparently a lot of people did not pay attention in grade school math and because 4 was bigger than 3, they thought 1/4lb was bigger than 1/3lb.
Numbers confuse the heck out of many people. So does electricity. Manufacturers use this to their advantage. The water analogy at the beginning is best.
Advertisers are well aware that a large proportion of their target audience is numerically illiterate and take advantage by using confusing terms. 50% off is a favourite.
I watch a number of quiz shows on TV and contestants often pass immediately if they hear a question that involves arithmetic. Even simple things like eight nines, or 30% of 90, are instantly passed. Questions like “What’s the name of [obscure celebrity’s] eldest daughter,”, or " Who lost the 1995 World Cup final," are deemed easier.
It’s the same psychology as pricing goods with a 99¢/99p to make them seem cheaper.
Here’s my from-the-ground-up primer on parameters used to describe properties of electrical systems:
Electrical charge is measured in coulombs. an electron carries an electrical charge of 1.06x10-19 coulombs.
Current (the flow of electrons in a conductor) is measured in amperes (amps). One amp equals 6.24x1018 electrons flowing past your measurement point every second. In other words, one coulomb of electrical charge flowing past every second.
Voltage describes the amount of energy carried by a unit of electrical charge. Voltage is measured in units of joules per coulomb. So if an electrical current experiences a drop of 1 volt when it flows through an electrical component, then every coulomb of charge that passes by delivers 1 joule of energy to that electrical component. If an electrical current experiences a rise of 1 volt when it flows through an electrical component (e.g. a battery), then every coulomb of charge that passes by receives 1 joule of energy from that electrical component.
Power is energy delivered per unit of time. For electrical devices, the usual unit of power is watts, which is joules per second. (For mechanical devices like motors, they are sometimes rated in horsepower. One horsepower is about 746 watts.)
So if you want to calculate power, you multiply current with voltage:
P = I * V
Using the units for each of those things, you can see why the units cancel out and leave you with the above tidy equation:
Power (joules/second) = current (coulombs/second) * volts (joules per coulomb)
Power (joules/second) = current (coulombs/electron * electrons/second) * (joules per coulomb)
If you want to fully understand an electrical device, you’d want to know all of those things - current usage, voltage requirement, and total power consumption. For things that plug into the wall in the US, the voltage is generally understood to be 120V, with an AC frequency of 60 Hertz (Hz, or cycles per second), but you’ll still want to know the total wattage or current usage.
For battery-powered things, you’ll also want to know the voltage. For a given power output, higher voltage means lower current flow. High current flows result in resistive losses in the wiring (and in the battery itself), which means less of the battery’s total energy goes toward making your device run. All other things being equal, a 10-volt cordless drill will probably not run as long (or be as powerful) as a 20-volt cordless drill. This is why electric cars have battery packs that operate at hundreds of volts: if you need 200 kw of power, a 400-volt battery pack can deliver this with just 500 amps of current, and reasonably-sized power cables. Try to deliver 200 kw with a 12-volt battery, and you’ll be drawing 16,600 amps, and you won’t be able to fit the cables in your car.
If the package really said that, then it’s nonsensical. 13 watts is a legit rating of electrical power consumption, expressing joules of energy used per hour. Adding “per hour” on the end is like reporting your boat’s speed in knots per hour.
Here is a graphic showing the relationships between power (W), voltage (E), current (I), and resistance (R), from this site:
The little circles on the bottom are a calculator for resistance, current, and volts; you cover the one you are looking for to figure what to math you need to do to the two you have, as shown on the right.
Yes voltage is not power. but a 40 volt battery can deliver more power for the same amps as a 6 volt without having to increase the amps, size of the wire, or motor. True battery voltage is not a true example of the power but an indication.
However there is no such thing as a free lunch. The energy content of batteries of the same size is the same. The 40 volt battery just has more cells of smaller size. The overall chemical energy available is a wash.
The actual power delivery is dependant upon the motor controller in the tool. Brushless motors can haul silly current, and therein lies the devil in the detail. Overall the power a motor can deliver is determined by the physical size and thermal constraints of the motor. The magnetic circuits don’t care how they were created, saturation of the circuit limits the power. Iron losses remain. They are proportional to the power of the motor.
There are some wins. For a higher voltage you can reduce the diameter of the winding wire and increase the number of windings. With the higher voltage you are thus able to reduce I^2R (aka copper) losses in the coils. This is where there is a win.
Motors get hot, and a small brushless motor is limited in capability by how hot it gets. By reducing the copper losses a motor of a given power heats up less. Or, for the same temperature rise, you can haul a bit more power out of it. But this gain is nothing like the V^2R power rise.
Pulling numbers out of the air. Say a motor is about 75% efficient. So say half of the losses are copper losses. Most of the remainder iron losses. Doubling the voltage might reduce the copper losses by half. So there is the potential to get up to say a 30% power increase before the motor overheats. I doubt they get all of that, but I would bet that this is the biggest win. (Numbers here are almost certainly bogus, but just indicative of the idea.)
Losses in the motor controller electronics are proportionately reduced as well. However modern switching MOSFETS and the like are remarkable in the small forward drop. But still a small win for higher voltages.
Battery cells cannot deliver arbitrary current. They all have an internal resistance, which is a combination of the construction and the nature of the chemistry. A good estimate of relative internal resistance between cells of similar chemistry and construction is the surface area of the electrodes. Which since battery cells are built by wrapping long layered foils is pretty much linearly related to the volume of the battery. The internal resistance of a 40v battery will be roughly at bit more than double that of the 18v battery. At the limit the current delivery is the much same for each. So a win for 40v.
Overall, there are useful gains in a 40v tool. But they are nothing like the gains a simple V \times I or {V^2}\over R would suggest.
One parameter of the Volt, Amp, Watt trio is pretty meaningless. But may sound good with a bigger number.
A 40 Volt battery versus a 20 Volt battery, does not tell you how much power it can deliver. You need to know how many Amp Hours it contains and how many Amps it can deliver continuously. Volts times Amps equals power in Watts. A battery can only deliver a certain amount of Amps at any moment without overheating and being damaged, or just has a maximum that it can deliver.
It is important to know the Amp or Milli-amp hours the battery can deliver. 40 Volt 10 Amp hour battery has about twice the run time of a 40 Volt 5 Amp hour battery. The Voltage is very important due to the device being powered by the battery, will be designed to work in that voltage range. There will also be a minimum Amp requirement too. Once you meet the minimum Amperage to operate at all, more Amp Hours equals longer run time.
So when choosing a battery. Volts is most important to be compatible with the device. Then, more Amp Hours means more run time. The battery also needs to be able to produce the minimum Amps to run the device, without damaging the battery.
Then comes the engineering of the device you are powering with the battery. A lot of battery power can be wasted, driving a poorly designed device. Waste heat, friction, inefficient design in many ways.
You can buy a cheap tool that comes with a battery that provides a lot of power. But does not run as long as you might expect.
Ugh. Mistake in what I wrote above. Twice as many cells and twice internal resistance each. So four times the internal resistance for a 40 V battery. So power delivery, at the limit of the battery’s capability, is actually the same for both batteries.
That is what I was trying to get to. (I knew the result, I couldn’t work out why I couldn’t get to it. Time for a coffee.)
I’m trying to figure out what the confusing part of “50% off” is, or how it’s supposed to be tricky or fooling the consumer in some way. That’s a hefty discount. I’m probably misunderstanding, though.
(I reformatted the above slightly because MathJax formatting was screwing up in the quote. I think I got it right.)
There’s a very simple concept in that first line that I think has been implied several times in this thread but, at the risk of oversimplifying, is worth calling out explicitly. It’s easy to get deeply into the weeds here technically, but this is really simple and basic.
Is something like a 20-volt cordless drill going to give you more power than, say, a 12-volt one? Technically, not necessarily, but you can see from that equation that in order to give you the same power, the battery is going to have to deliver more current (amps), and more current is going to have to run through the internal wiring and the motor. So for practical reasons, yes, the 20-volt drill in the real world will always be more powerful than the 12-volt. But in theory you could build a very powerful 12-volt drill. After all, a 12-volt battery operates your car’s starter, which must produce a tremendous amount of power to turn over that big engine. But the amperage required to do so is huge.
Fun fact: cars running on the old 6-volt system instead of a modern 12-volt would need heavier cabling for any given amount of power delivery, especially to high-power devices like the starter. This seems counterintuitive, but the reason is that the wires have to carry twice the current (amperage).
When dealing with high-power systems like subway trains, a typical operating voltage is about 600 volts. The high voltage keeps the amperage requirements from becoming ridiculously high. It also helps to minimize voltage drop across long distances (because voltage is analogous to “pressure”). OTOH, you don’t want to use too high a voltage because then you have insulation issues.
Also just to add, on that second point, Joules is the common scientific measure of energy, but another everyday unit is watt-hours (power over time), or as commonly seen on your electric bill, kilowatt-hours (KWh).
Partly because they don’t say ‘off what’. The real price or some imaginary price made up for the sale? Is it before sales taxes or after?
I am sure that almost all the readers of this forum are well aware that 50% is a half, but many others have no idea at all. They do expect (rightly) that 60% off is better than 50%, but what about 20% extra off? Is 50% plus an extra 20% the same as 70%?
I picked 50 as a possibly bad example. Ask the ubiquitous person on the Clapham Omnibus what 20% of 100 is and most of them will not know.
This also explains why we in the UK with our 230V power supply can use thinner wires in our houses than you guys in the USA.
I’ve never seen the basis being anything other than the original selling price before the “50% off” was applied". To do otherwise would, I think, amount to false advertising.
Sales tax is calculated last, after all other fees and discounts have been applied. So if a $10 item has 60 cents of sales tax added on it (total, $10.60), then when it’s 50% off, it’ll be $5 with 30 cents of sales tax on it (total, $5.30). In the end, it’s 50% off no matter how you look at it.
Aircraft use 400v electrical systems primarily to lower the weight of the equipment and cabling.
Plus of course, the ubiquitous,
As much as 50%Off!
or
50%Off select items
YOu ever seen a store going out of business sale. 1st step is everything is priced at 50% off. What they don’t tell you is it is not 50% off the price of the merchandice the week before the going out of business sale. But it is 50% of the MSP. So a shirt in a store that had a MSP of 30 dollars when 1st put on t he sales floor may have been reduced to 20 dollars. Then after it sat there for a month it was reduced to 15 dollars. and the next automatic sales reduction put the price at 10 dollars. Next week the stor goes out of business and the 50% off price it 50% of 30 so it is sold for 15 dollars. I saw a lot of that kind of pricing when working for a department store chain that started closing stores, and then going out of business.
You are correct when you ask 50% off of what?