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You seem like a pretty angry guy there. I think you might find what you seek if you kept your post less on the confrontational side, less wordy and keep to simple questions. Like I never saw you ask:

Which is quite simple.no one ever did answer my question about how amps correlate to amp HOURS or that other time based roadblock to using the watts per velocity & mass formula.

Ampere hours (Ah) is a unit of charge. Ampere (A) is a unit of current or defined as a rate of charge. Therefore, charge equals current times time. Or Ah = A × h. So if you have 2 A flowing for 3 hours, it equals 6 Ah of charge. 6Ah = 2A × 3h.

This is a little difficult for me to understand what you are looking for. But velocity (v) and mass (M) relate to energy in the form of kinetic energy (KE) or energy associated with motion, by the formula KE = ½Mv².time based roadblock to using the watts per velocity & mass formula

Watts (W) are the units for power. We all know what time is and the units are seconds (s). Any type of energy has the units of joules (j) which are defined as watt seconds (Ws). So KE = Ws.

Therefore your quest to find "watts per velocity & mass" is W / (½Mv²) = W / Ws = 1/s. So you are correct that time (or the inverse) is watts per velocity and mass, which for some reason you call a roadblock. But I guess time is the roadblock of the universe.

Regards,

major

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7,793 Posts

Hey zero,

Do you ever read replies to your posts?

major

after wading through the nonsense to get to the allegedly CORRECT formula for how many kj are needed to push 2135 lb @ 55mph, the math becomes USELESS anyways because another online conversion utility says that 300kj (rounded up from 292) = just 83w!!! there's no way that can be right.

velocity (v) and mass (M) relate to energy in the form of kinetic energy (KE) or energy associated with motion, by the formula KE = ½Mv².

Watts (W) are the units for power. We all know what time is and the units are seconds (s). Any type of energy has the units of joules (j) which are defined as watt seconds (Ws). So KE = Ws.

One Watt of power is equivalent to one Joule of energy expenditure per second. It may require 300 kJ of energy to accelerate a car to 55 mph but that would generally depend on the amount of time taken to do the acceleration. Once you're moving, maintaining a given speed is dependent on your friction and drag losses, and requires a certain rate of energy expenditure - i.e. a certain amount of power.

---------------------------------1 watt-s = 1 joule. Also previously stated, a watt is instantaneous work. Add the time factor and it becomes an amount of energy.

300,000 joules = 300,000 watt-s, or 5000 watt-minutes, or 83 watt-hours.

If you were to provide 83 watts of power over the period of an hour, you'd be using the energy equivalent of 300,000 joules.

If you used a more useful period, say the acceleration time of a car going from 0-55 mph over 10 seconds, you'd have 50,000 watts over 10 seconds. That's 50,000 watts delivered for 10 seconds equaling the energy of 300,000 joules... ~65hp for 10 seconds. Reasonable for a 2000lb car.

It's all there. People haven't given you the specific answer to your numerical problem, but they've given you the tools to figure it out.

You are correct. This is incorrect. kj cannot equal w. Energy and power are different quantities. I suggest you brush up on some basic physics fundamentals and then it will become easier to correctly apply the math.another online conversion utility says that 300kj (rounded up from 292) = just 83w!!! there's no way that can be right

Regards,

major

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7,793 Posts

We're not being difficult. Do you read the replies to your posts?why does EVERYONE in this forum have to be so difficult?

We have posted regarding this several times. I will put it as simple as I can. MASS times VELOCITY equals NOTHING !!!!!!!!!!! Meaning that the product of mass and velocity does not result in any useful quantity in any known system of physics or units in this universe, as far as I am aware.*why on earth should i have to study physics when all i want is a simple formula (or group thereof) that converts mass & velocity into watts? that's ALL i'm looking for eg. mass x velocity x .1234 = ergs, ergs divided by 4.321 = horsepower & horsepower x 9.876 = watts. that's all i'm trying to get to, but can't find the COMPLETE formulas. if KJ can't be converted to watts, then that's the wrong formula.

Furthermore, an erg = 100 nanojoules, which is a unit of energy. ENERGY is not POWER. So you can NOT convert units of energy to units of power.

No simple factor (number) can be used to make a KJ into a watt.

Got that?

major

* edit. Mass times velocity equals momentum. I missed that. Not a quantity used much in the EV study, but nevertheless, a useful quantity in this universe afterall.

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7,793 Posts

Good point. I stand corrected. Momentum. Not something I use much. I double checked several places before posting that and could not see where the unit of kg*m/s was listed anywhere. Also did not see N*s. Oh wellHey Major,

I'm not trying to quibble here but Mass times velocity equals momentum.

p = mv

But I do read replies to my posts

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7,793 Posts

My God. I think you have read a reply In that case maybe the following will help.the round peg in a square hole analogy is very appropriate for this thread.

Please read the EV Information section of this site. http://www.diyelectriccar.com/forums/showthread.php?t=669now if only i could find the formulas for moving that object at 55mph in relation to planet earth i'd use the word progress.

There, under this sub topic, you will find http://www.diyelectriccar.com/forums/showthread.php?t=15508

And in that you can read this:

1. The POWER requirements of your car at a particular speed is:

Power in Watts = ((Mass in kg) (9.8m/s²) (Velocity in m/s) (Rolling Resistance)) + ((0.6465) (Coefficient of Drag) (Area in m²) (Velocity^3))

The 9.8m/s is acceleration due to gravity, the 0.6465 is 0.5 times the density of air in kg/m^3. If you enter the required numbers it will give you your power consumption in watts.

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