# Boost by Nation



## Zipper730 (Feb 5, 2020)

The US used Inches of Mercury, the UK used PSI relative to atmospheric, the Germans used ATA's (I'm not sure the conversion formula exactly), and the Japanese used millimeters of Mercury relative to atmospheric (1 ATM = 0.00)?


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## AMCKen (Feb 6, 2020)

I made up a formula to use in basic.
pounds = ((inches - 29.9213) / 29.9213) * 14.69595
1 ATA / atm is 29.9213 inHg or 14.69595 psi

35 inches is 2.5psi and 1.17ata (atmosphere absolute)
AFAIK.

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## Zipper730 (Feb 6, 2020)

A
 AMCKen

W
 wuzak


The USSR seemed to use mm H2O. I'm not sure if this was absolute or gauge pressure.


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## mikewint (Feb 6, 2020)

I can't see using mm of water to measure pressures unless we're talking about very small pressures, One Atmosphere of pressure will support a column of water roughly 33 feet tall (10.06m). Possibly you read mm of *Hg* where Hg is the chemical symbol for Mercury or you added one extra *m* so mH2O for Meters of water thus 1 atm = 10.33 mH2O
So to relate: 1 Atmosphere = 101325 Pa 1 Bar = 100,000 Pa and the German ATA (technical atmosphere absolute) is Kilograms per centimeter squared
so - 1 ATA = 98066.5 Pa
On a clear day at sea-level under ideal circumstances engine turned off a manifold gauge would read: 101325 Pa - 0 psi - 29.92 in of Hg - 1.0332 ATA

In passing: one standard atmosphere will support a column of Mercury 760 mm tall thus 1 ATM = 760 mm of Hg. Pre SI, mm of Hg were called TORR
Since Mercury is 13.6 times denser than water 760 X 13.6 = 10336 mm or 10.336 meters


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## Zipper730 (Feb 6, 2020)

mikewint said:


> I can't see using mm of water to measure pressures unless we're talking about very small pressures


I'm not sure why they'd use mm H2O either. That said, it seems bizarre to use +PSI as you end up with negative numbers.


> So to relate: 1 Atmosphere = 101325 Pa 1 Bar = 100,000 Pa and the German ATA (technical atmosphere absolute) is Kilograms per centimeter squared
> so - 1 ATA = 98066.5 Pa


So that's how they came up with ATA's... that makes sense.

I'm just curious for the calibrations used for the performance charts, if they used 0.980665 Bar or some rounding figure. For example, I'm not sure if for the performance charts, if they used 14.696 or 14.69595, and if they used 29.92 solely or 29.9213. With Excel I can enter any of these numbers, and they'll compute out.


> In passing: one standard atmosphere will support a column of Mercury 760 mm tall thus 1 ATM = 760 mm of Hg.


That's quite a lot of mercury


> Since Mercury is 13.6 times denser than water 760 X 13.6 = 10336 mm or 10.336 meters


That does make using water a bit strange...


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## Zipper730 (Feb 7, 2020)

W
 wuzak


 mikewint


I'm curious the conversion for PS to HP and HP to kW? I largely have a viable chart at this point.


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## wuzak (Feb 7, 2020)

You can't find that yourself?


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## Zipper730 (Feb 7, 2020)

wuzak said:


> You can't find that yourself?


The problem is I'm not sure what calibrations were used at the time to create the figures. For example 14.70 is commonly seen, it seemed the RAF used 14.696. Technically 14.69595 can be used, but I'm not sure if they entered those numbers, and I'm figuring it might have a slight effect that wasn't in the actual documents.


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## wuzak (Feb 7, 2020)

The first and second are rounded from the third.


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## Zipper730 (Feb 7, 2020)

wuzak said:


> The first and second are rounded from the third.


So, I'll just put 14.69595. Looking at the workbook, the numbers come out close enough.


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## mikewint (Feb 7, 2020)

First of all POWER – defined as the RATE at which work is done and WORK is FORCE X DISTANCE.

**N.B.** The Force exerted must produce a displacement so strangely enough simply holding a 20 pound weight stationary 4 feet above the floor *does zero work*! Raising a 20 pound weight to a height of 4 feet above the floor and then lowering it back down *does zero total work *since the positive lifting work and the_ negative lowering work cancel. R_aising the weight from the floor to a height of 4 feet does a positive work. 20lb X 4ft = +80 ft x lb of work

PS: PferdeStärke is the standard unit used in the EU (European Union) to measure the power of an engine DIN (Deutsches Institut für Normung) 66036 defines one *metric horsepower* as the power to raise a mass of 75 kilograms against the Earth's gravitational force over a distance of one meter in one second: 75 kg × 9.80665 m/s2 × 1 m / 1 s = 75 kg⋅m/s = 1 PS. This is equivalent to 735.499 W, or 98.6% of an imperial mechanical horsepower.

James Watt did not invent the steam engine. He did however improve the efficiency of the Newcomen steam engines by a factor of Five giving a tremendous savings in coal usage by the engine. In the early 1700s work was done by horses and coal was the main source of energy. Coal mines tended to fill with ground water which had to be constantly pumped out of the mine shafts. The pumps were turned by teams of horses which needed to be rested and replaced on a regular basis. So a single pump would require 10 – 12 horses to keep it operating continuously. To sell his engines then Watt need to show that one of his engines could do the work of 10 – 12 horses. Thus Watt determined that a horse could turn a mill wheel 144 times in an hour (or 2.4 times a minute). The wheel was 12 feet (3.7 m) in radius; therefore, the horse traveled 2.4 × 2π × 12 feet in one minute. Watt judged that the horse could pull with a force of 180 pounds-force (800 N). So:

P = W / t = F x d / t = 180 lb X 2.4 X 2pi X 12 ft / 1min = 32,572 ft x lb  / min

Thus Watt defined the horsepower as 32,572 ft⋅lb/min, which was rounded to an even 33,000 ft⋅lb/min.

The WATT is the SI unit of Power defined as a Joule per sec = a Newton x meter per second = a Kilogram x Meter^2 per sec^3. Converting to equivalent English units a Horsepower = 745.7 Watts

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## tomo pauk (Feb 7, 2020)

There you go.

BTW (edited):


Zipper730 said:


> The USSR seemed to use mm H2O Hg. I'm not sure if this was absolute or gauge pressure.



Absolute.


Zipper730 said:


> I'm not sure why they'd use mm H2O either.



French and Italians used it, too (mm Hg).


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## Zipper730 (Feb 7, 2020)

345.32508 mm H20 is atmospheric pressure?


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## tomo pauk (Feb 7, 2020)

I must check 1st.

Sorry for posting wrong data. Soviets used "mm Hg", not "mm H2O". In cyrilic, it is 'mm рт. ст.'; the ''рт' is 'ртуть' = 'mercury', 'ст' is 'столбец' = 'column'.


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## pbehn (Feb 7, 2020)

mikewint said:


> I can't see using mm of water to measure pressures unless we're talking about very small pressures, One Atmosphere of pressure will support a column of water roughly 33 feet tall (10.06m).


 Yup, we did that experiment at school when I was 13.


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## wuzak (Feb 7, 2020)

mikewint said:


> First of all POWER – defined as the RATE at which work is done and WORK is FORCE X DISTANCE.
> 
> **N.B.** The Force must be exerted *against gravity* so strangely enough simply holding a 20 pound weight stationary 4 feet above the floor *does zero work*! Holding a 20 pound weight at a constant 4 feet above the floor as you move laterally 20 feet also *does zero work.* Now raising the weight from the floor to a height of 4 feet does do work. 20lb X 4ft = 80 ft x lb of work



The Force does *NOT* have to be exerted *against gravity*.




> n physics, work is the product of force and displacement. A force is said to do work if, when acting, there is a movement of the point of application in the direction of the force.
> 
> For example, when a ball is held above the ground and then dropped, the work done on the ball as it falls is equal to the weight of the ball (a force) multiplied by the distance to the ground (a displacement). When the force {\displaystyle F}F is constant and the angle between the force and the displacement {\displaystyle s}s is θ, then the work done is given by W = Fs cos θ.



Work (physics) - Wikipedia

Work is a vector quantity, so if you spend a lot of energy moving the couch all around the lounge room only for it to end up in the same spot you've done no work.


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## Shortround6 (Feb 7, 2020)

wuzak said:


> Work is a vector quantity, so if you spend a lot of energy moving the couch all around the lounge room only for it to end up in the same spot you've done no work.



DO *NOT* give my wife any ideas.

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## mikewint (Feb 8, 2020)

wuzak said:


> The Force does *NOT* have to be exerted *against gravity*.


You are correct and I stand corrected. I was thinking in terms of gravitational potential energy. To do work there must be a displacement in the direction of the force. Pushing all day against a building until you collapse does zero work and although Force was exerted there was no displacement and thus no work was done



wuzak said:


> you've done no work.


Add the word "TOTAL" and we're in agreement pushing the couch and moving it does do work. Eastward displacement does positive work as does moving it northward. Westward movement and southward movements are negative displacements. If the couch returns to the same location the the vector sums will indeed total zero


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## ThomasP (Feb 9, 2020)

Technically, by the physics definition of work, the energy state of the object/system on which the work is performed must have a net change in energy state or 0 net work has been done. Displacement, or a change in location, of an object/system will only count as work if it also changes the energy state of the object/system.

As Mike stated above, if you raise an object 4 ft above the floor you do work, ie you have increased its potential and kinetic energy. If you then lower it back to the floor you do work, ie you have decreased its potential and kinetic energy by the same amount. The increase and decrease in energy results in 0 net work.

However, unless I am misunderstanding what Mike and Wuzak were saying, it needs to be said that the couch does not have to end up in the same place as it started. If the couch started at 0 mph relative velocity and at 0 ft relative altitude, it does not matter if it was moved sideways...as long as the velocity and altitude change are 0 relative, then the energy state has not been changed and 0 net work has been done. (I am assuming we are not counting minute irregularities in altitude of the floor or minute changes in latitude.) Another way of saying this is, imagine if the couch were on an ice rink, and Shortround's wife says "it would look better over there". When Shortround dutifully gives it a push in the proper direction he is accelerating the couch thereby increasing its relative kinetic energy. If Shortround then moves to the other side and pushes in the opposite direction (or just lets it slow down due to the effects of friction and cohesion) until it is again stationary, the change in relative kinetic energy back to 0 will result in 0 net work.

When you change rotational velocity, temperature, electric & magnetic field strength, strong force, weak force, etc, you are also doing work, although the work done is often defined differently than in the our examples above.


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## mikewint (Feb 9, 2020)

ThomasP said:


> it does not matter if it was moved sideways.


It matters. If you apply a Force and that Force or component thereof produces a displacement in the direction of that force or component then you do work.
Now if the floor is perfectly level then any lateral movement does not change the gravitational potential energy of the object thus zero work is done against the gravitational force (assuming that the distance moved is small and no change in g occurred) and no matter where the object ends up the gravitational potential energy remain unchanged. Now assume several rises and dips in the floor. Lateral movement over the rises ingrease PEg so positive work must be done and as the object traverses a dip PEg is decreased as the gravitational force does negative work. So depending upon where the object ends up the total PEg may be +, -, or 0.
Back to the level floor. North and East are positive displacements while South and West are negative. Move the object where you will and then sum the N & S displacements from the starting point (0,0) to see if there has been a net + or - displacement along the Y-axis. Then sum the E & W displacements to find the net + or - displacement along the X-axis. If both total zero then the object is back at the origin and zero net work has been done.
However PEg has not changed and all the work has been done against the frictional force. The energy expended has been converted into thermal energy and the floor, object, and room air are a bit warmer as a result.
If you want to play around with trig we can do angular displacements such as traveling 10 m due west followed by 17.32 m due north or conversely traveling 20 m 30 degrees W of N (bearing 330 degrees if you prefer)


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## Shortround6 (Feb 9, 2020)

Now my head hurts in addition to my back.

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## pbehn (Feb 9, 2020)

Shortround6 said:


> Now my head hurts in addition to my back.


I think the discussion is working around in circles.

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## ThomasP (Feb 10, 2020)

Hey Mike,

The answer is in the math.

In the example of the couch above, lets say that Shortround pushes the couch from V0 at X0 over a distance D1 in the positive X direction, using a force F1 resulting in a velocity V1. Shortround then moves (instantly) to the opposite side of the couch and pushes in the opposite direction using a force F2 over a distance D2 in the positive X direction until the couch comes to a velocity of V0 at X2. Since the Work (W1) done to accelerate the couch to V1 has to be equal to the Work (W2) done to decelerate the couch to a velocity of V0, the net work is 0 even though the couch is displaced by a distance of D1 + D2.

This is because the energy state of the couch is the same at X2 as it was at X0.

You can substitute any values in the above formula as long as the result is a velocity at X0 that equals the velocity at X2.


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## Zipper730 (Feb 10, 2020)

Okay, let's get back to it. I'll put up my chart, and you guys tell me if you like the way it looks, if all the numbers are filled out decently.


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## Zipper730 (Feb 10, 2020)

BTW: How do you switch axes in Excel graphs?


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## wuzak (Feb 10, 2020)

I think the mmHg would probably be absolute pressure. It would depend on teh country, though.

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## wuzak (Feb 10, 2020)

ThomasP said:


> Hey Mike,
> 
> The answer is in the math.
> 
> ...



Your logic means there would never be any work done where the couch has the same final velocity and altitude. 

Work is simply the force multiplied by the distance. The total work in your example is W = F1.D1 + F2.D2.

W=0 only when F1.D1 = -F2.D2

The thing missing from your calculation is the time between the acceleration phase and the deceleration phase, where the velocity and force remain constant.

Also, in the couch example, friction works against you for acceleration and with you for deceleration. Not often when moving furniture do you have to run to the other side and push to slow it down.


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## Zipper730 (Feb 10, 2020)

wuzak said:


> I think the mmHg would probably be absolute pressure. It would depend on teh country, though.


I'm personally a fan of absolute pressures myself, but the Japanese used gauge pressure in mmHg.


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## Zipper730 (Feb 10, 2020)

When it comes to making graphs on Excel. I've figured out how to alter the axes for climb & altitude but when I select altitude and climb-speed, I get some very funny numbers.

The altitude section looks normal but the speed figures are like 0-20. Instead of 151.5 being the lowest climb-speed listed.


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## mikewint (Feb 10, 2020)

ThomasP said:


> The answer is in the math.


It is indeed _except_ that you are equating different types of work. On a flat level friction-less surface W is simply F (force applied) X D (displacement)
In your scenario you are including the force needed to produce an acceleration to a constant velocity. In this case W = *Δ* KE. Considering that the couch was initially at rest, initial KEi = 0 thus Wt = (0.5)mv^2. On a frictionless surface the couch will continue to move at a constant velocity unless an opposing force is applied. When that opposing force is applied and it is equal and opposite to the initial force then the couch will come to rest. So the Work done to accelerate and the work done to decelerate (net Work) do indeed total zero in this frictionless pretend world but in the real world friction has to be contended with and a constant force must be applied to maintain that constant motion. That "extra" force is in the direction of motion and there has been a change in d (displacement) so that work remains unchanged since W still = F d


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## pbehn (Feb 10, 2020)

It all depends what you want to know, a plough ends up back in the same place it started at after a day ploughing a field, that doesn't mean that no work has been done.


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## mikewint (Feb 10, 2020)

pbehn said:


> that doesn't mean that no work has been done.


Depends on how you define "WORK". In Physics WORK (one definition) is defined as FORCE x DISPLACEMENT and the displacement must be in the same direction as the force applied AND WORK is a vector quantity so a displacement North followed by an equal displacement South are adding a + number to a - number which totals Zero like it or not. So NO *NET *Work was done. Positive work was done as the plow went North and Negative Work was done as the plow moved South. So individual quantities of Work were indeed done but the *NET TOTAL *Work remains zero. By the same token you can hold up a weight until your arm turns black and falls off but NO Displacement means NO Work was done. Push on your house until you collapse in a puddle of sweat. No displacement NO Work.
Consider Post #23. WORK is also defined as WORK = Δ KE or Work(t) = KEi + KEf. IF we live in a frictionless world then object in motion remain in motion. Thus we can apply an instantaneous force to accelerate to a velocity and the object moves at constant velocity with no extra force applied. There has been a change in KE thus Work was done. At some point in time apply a second instantaneous equal and opposite force and the object stops. There has been a second change in KE. BUT that first KE change was positive and the second was negative thus WORK TOTAL = Zero and the weird part is that during the constant velocity phase a displacement occurred BUT BUT BUT even though Work = F d NO WORK was done to produce this d since the F(net) was zero during this constant velocity phase. *PHYSICS IS PHUN!!! *


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## pbehn (Feb 10, 2020)

mikewint said:


> Depends on how you define "WORK". In Physics WORK (one definition) is defined as FORCE x DISPLACEMENT and the displacement must be in the same direction as the force applied AND WORK is a vector quantity so a displacement North followed by an equal displacement South are adding a + number to a - number which totals Zero like it or not. So NO *NET *Work was done. Positive work was done as the plow went North and Negative Work was done as the plow moved South. So individual quantities of Work were indeed done but the *NET TOTAL *Work remains zero. By the same token you can hold up a weight until your arm turns black and falls off but NO Displacement means NO Work was done. Push on your house until you collapse in a puddle of sweat. No displacement NO Work.
> Consider Post #23. WORK is also defined as WORK = Δ KE or Work(t) = KEi + KEf. IF we live in a frictionless world then object in motion remain in motion. Thus we can apply an instantaneous force to accelerate to a velocity and the object moves at constant velocity with no extra force applied. There has been a change in KE thus Work was done. At some point in time apply a second instantaneous equal and opposite force and the object stops. There has been a second change in KE. BUT that first KE change was positive and the second was negative thus WORK TOTAL = Zero and the weird part is that during the constant velocity phase a displacement occurred BUT BUT BUT even though Work = F d NO WORK was done to produce this d since the F(net) was zero during this constant velocity phase. *PHYSICS IS PHUN!!! *


That is what I said.


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## Joe Broady (Feb 11, 2020)

Zipper730 said:


> The problem is I'm not sure what calibrations were used at the time to create the figures. For example 14.70 is commonly seen, it seemed the RAF used 14.696. Technically 14.69595 can be used, but I'm not sure if they entered those numbers, and I'm figuring it might have a slight effect that wasn't in the actual documents.



Do you realize that 14.7 psi, which you probably learned as a schoolboy, is within 0.03% of the standard sea level pressure of 1.01325 bar? That's an error of 1 part in 3600. I think you're agonizing over trifling differences between numbers. Isn't 0.03% accurate enough for what you're doing?

A sign of a numerate person is precision which is sufficient but not excessive.

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## ThomasP (Feb 11, 2020)

Hey Wuzak,

re your post#27: "Your logic means there would never be any work done where the couch has the same final velocity and altitude."

Please reread my post#23. My statement is that there was 0 net work. You can add in additional F3 x D3 (ie friction), F4 x D4 (ie cohesion), F5 x D5 (ie wind resistance), Fn x Dn, etc .... it will not change the value of net work if the velocity at Xn ends up equal to 0.


Hey Mike and Wuzak,

I think there may be a slight misunderstanding here as to what work actually is. Work (physics definition) is defined as doing something to an object or system that changes its energy state.

In a mechanical system this is usually defined as changing the energy state of the object/system by applying a force over a distance (ie F x D) which would change its kinetic energy (KE), which is why W=ΔKE. Since Force is equal to Mass x Acceleration (ie F=M x A), in order to have net work (ie ΔKE) you would have to have accelerated the object/system to a velocity not equal to the start velocity.

In Mike's post#11 he states: "Raising a 20 pound weight to a height of 4 feet above the floor and then lowering it back down *does zero total work *since the positive lifting work and the_ negative lowering work cancel."_ The reason that this true is that the work done by raising the 20 pound weight 4 feet raised the Potential Energy (PE) of the object and lowering the object reduced the Potential Energy of the object by the same amount. In this case W=ΔPE, and since the PE ended up the same after the object was lowered back to its start height there was 0 net work. It should be noted that this would also be true if the object was raised 4 feet, moved sideways 4 feet, and then lowered 4 feet back to the floor, ie ΔPE=0

As I stated in my post#19, work may be described in equivalences, ie mechanical (ΔKE), heat (ΔT), magnetic (ΔG), etc. In none of the equivalences will you find the result of work listed as just feet, which is what you would have to be able to do if displacement without a net change in energy state counted as work. I am not trying to be snarky here but go ahead and try to find any variation of work done, net or otherwise, described as
"= x feet" or "Δft".


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## wuzak (Feb 11, 2020)

> [When a force acts upon an object to cause a displacement of the object, it is said that work was done upon the object. There are three key ingredients to work - force, displacement, and cause. In order for a force to qualify as having done work on an object, there must be a displacement and the force must cause the displacement. There are several good examples of work that can be observed in everyday life - a horse pulling a plow through the field, a father pushing a grocery cart down the aisle of a grocery store, a freshman lifting a backpack full of books upon her shoulder, a weightlifter lifting a barbell above his head, an Olympian launching the shot-put, etc. In each case described here there is a force exerted upon an object to cause that object to be displaced./QUOTE]
> 
> Definition and Mathematics of Work
> 
> ...


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## mikewint (Feb 11, 2020)

ThomasP said:


> any variation of work done, net or otherwise, described as
> "= x feet" or "Δft".


In Ye Olde Golden Days of Yore Physics problems were worked in one of three coherent systems of measurement. Either the MKS or the CGS or the FPS systems, MKS - Meter Kilogram Second; CGS - Centimeter Gram Second; or the English FPS - Foot Pound (unit of Force) Second.
So the Work formula W = F x d could have 3 answers depending on the system used, i.e. MKS - 20 Newtons x 10 meters = 200 Nm or Newton Meters; CGS - 20 dynes ( 1 g⋅cm/s^2) x 10 centimeters = 200 ergs (1 g⋅cm^2/s^2); and finally FPS - 20 pounds x 10 feet = 200 ft⋅lb. 
In the US strangely torque wrenches are calibrated in inch-pounds or foot-pounds(ft-lb) (the torque produced by a one pound weight hanging at the end of a 1 foot long lever)
With the almost universal switch to the SI system today Physics problems are always couched in the MKS system.
Historically the very term "WORK" was introduced by a French mathematician in 1826 and defined as "weight lifted through a height". The usage originated from measuring the output of early steam engines lifting buckets of water from flooded mine shafts. In the SI system the work unit is the JOULE which is defined as one newton of force acting through one meter of displacement. To make things even funner: this would be a Newton-meter but that is also a Torque unit which is also defined as force x distance or N-m. The Joule is also a unit of energy so WORK transfers ENERGY. Joule also measures HEAT energy since a Joule of Heat is produced by a current of one ampere passing through a resistance of one Ohm for one second.
WORK and KINETIC ENERGY are also related in that the WORK done by all forces acting on a particle equals the change in the particles KINETIC ENERGY.
Thus WORK can change the POTENTIAL ENERGY of a mechanical system; the THERMAL ENERGY of a Thermal system; the ELECTRICAL ENERGY in an Electrical system. In effect WORK transfers ENERGY from one place to another or from one form to another

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## AMCKen (Feb 13, 2020)

As mikewint said "In passing: one standard atmosphere will support a column of Mercury 760 mm tall thus 1 ATM = 760 mm of Hg. Pre SI, mm of Hg were called TORR "

TORR for Torricelli, the fellow who invented the barometer. 
Evangelista Torricelli - Wikipedia


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