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[Zipper730]

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)?

 

[AMCKen]

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.

 

[Zipper730]

A AMCKen W wuzak

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

 

[mikewint]

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

 

[Zipper730]

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...

 

[Zipper730]

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.

 

[wuzak]

You can't find that yourself?

 

[Zipper730]

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.

 

[wuzak]

The first and second are rounded from the third.

 

[Zipper730]

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.

 

[mikewint]

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. Raising 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

 

[tomo pauk]

There you go.

BTW (edited):

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

Absolute.

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

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

 

[Zipper730]

345.32508 mm H20 is atmospheric pressure?

 

[tomo pauk]

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'.

 

[pbehn]

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.

 

[wuzak]

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.

 

[Shortround6]

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.

 

[mikewint]

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

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

 

[ThomasP]

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.

 

[mikewint]

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)