where does the power go?

[Chris Smith]

When we added 100 amp [1200 watt] alternators to this size belt on other vehicles, they smoked the belt. <p>When we tightened the belts to their max deflection rate, [suggested by the manufacture] they snapped.

5 hp is impossible for any single wide belt of this size, which includes the 1/4 3/8 5/16 and ½ inch belt sizes. <p>Even the 3/4 belt wears out prematurely.<p>There never was a question of a mistake, 5HP and a VW belt are incompatible fihgures.

[terri]

I wish you'd said that before and saved me the embarrassment. <p>[ September 14, 2005: Message edited by: terri ]</p>

[Chris Smith]

Sorry, I just assumed it was a TYPO, and the others didn’t know?

[Robert Reed]

Proper automotive battery charging is done at a target voltage of 14.5 volts, not 12 volts. This puts the generator out put at 435 watts, not 360 watts, albeit a minor point.
With wood working machinery, I have seen 2 HP machines run with a single 1/2" v-belt and without slippage or premature wear. However most high end machinery will use a rule of thumb of one belt per horsepower.

[Chris Smith]

Its all about mass, acceleration and momentum. <p>Driving two horse power on one belt doesn’t mean two hp is in use at all times? <p>On a electric motor, there is no acceleration and deceleration going on constantly.<p> A rather smooth transition takes place, even on geared down lathes, or compressors except for the first moments of starting, or clutch engagement.<p>
On a car, the motor is at one moment idling at 500 RPMs, the next moment its at 5000 RPMs. <p>If the car is a sports car like a twin cam, the acceleration time frame is probably half that of the VW. <p>With the mass of the armature and the resistance of the charge, you actually have mass and momentum fighting the belt constantly, hence the burn out. Even if it doesn’t slip more than the standard 5 % to 10%, the belt gets hot enough to burn your hand, and thus it stretches and slips even more. <p>I have a half hp lathe that runs off a rubber band,... literally, a large “O” ring with no fiber in it like a fan belt.<p> IT works fine because other than start up, mass and momentum don’t come into play. <p>And most important is the diamater of the pulley.<p> Larger ones have more surface area of contact than the smaller ones with only a few inches of contact. As the belt size goes up, so does the surface area of contact. <p>I forget if its 33% or 50% in surface area of contact, when a belt size is increased from one to the next? <p>But also, the pulley is only designed for one or two sizes, so increasing size may not add surface area if the belt is incorrect.

[Will]

I like the ' doing calculations in the head part ' - According to my 'head' calculations - (which have equal probabilities of being right or being wrong) if the VW fan were of such a diameter (12 - 18 inches .?) that its' area were 1.0 sq ft. then, to absorb 4.5 hp the air would have to be accelerated by the fan to a velocity of something like 140 ft/sec - which is of the order of 850 mph - seems unlikely ?
i will get the calculator out tomorrow and check the 'head' calculation

[Will]

Whhops ! - I just poured some water over my head - I think I meant 130 ft/sec and 88 mph

[Chris Smith]

At that speed, you could street sweep with the air coming out!

[Will]

I guess we will just have to leave Chris with his 'pounds.ft' while we, and the rest of the world use 'ft.lbf' Something tells me that I went to school before Chris did and I was very clearly taught 'foot-pounds' then. I did find a reference on Yahoo which said that James Watt used the term 'pounds-ft' and I think Chris must have seen that one. But I found another that attributed Foot-Pounds to Watt. In any event, as to electrical units, the ft.lbf does not equate exactly to any electrical units. The earliest units which equated mechanical work/energy and elctrical units were 'ergs' or dyne.cms. . . and were introduced in 1783 (That might have been about the time Chris was going to mechanical school) The 'erg' (From a Greek word meaning 'to work' is that work done when a force of one dyne acts through a distance of 1.0 centimeter i.e. one dyne.cm. (I think 10,000,000 ergs = one Joule) For complitude the dyne is that force which will accelerate a mass of one gram at the rate of 1.0 cm/sec^2. Since, for (c.g.s) physical/mechanical work units, the force (dyne) term was placed before the linear (cm) term then this might seem to equate with 'Pounds.ft' It was changed when SI units were introduced and the ST unit for physical/mechanical work is the Newton.Metre (Which also has the force unit first) being equal to One Joule. One Newton being the force required to accelerate a mass of one kilogram at the rate of one metre/sec^2. It is possible, if one insists on using absolute definitions instead of derived relationships but I have always regarded that as a little stiff collared and impractical. That way one could define the Newton as 1.0 kg.m/sec^2 and, hence the Joule as one kg.m^2/sec^2 There is no accident here - these are precise relationships so either the Watt was derived from the Dyne.cm or the dyne was derived from the Watt. One couldn't suggest that the 'cm' came from any electrical units since it was originally (And still is with remarkable accuracy) one billionth part of the distance from the North Pole to the Equator on (Of course !) the meridan which runs through Paris - Not meridan zero which runs through London. _ I still derive some sort of comfort from thinking of a bunch of Frenchmen struggling with surveyors chains all the way from the North Pole to the Equator.
generators are not rated by the hour (Or by kWa) they are rated continuously i.e. in kW. If you insisted on using the term kWh in the rating of a generator (Unless you only expected it to run for only one hour) then you would have to divide by hours to get the continuous rating i.e. kWh/h = kW = kiloWatts.
You could not store a 'Watts worth' of energy (Because it is a POWER rating) unless you stated for how long you were storing it.

[Chris Smith]

Some of us passed our mechanical engineering class, by using the correct terms. <p>Others like todays students have lost the art of the proper term and even the English language, and make excuses for their bad education. <p>Oh well, my degree in mechanical engineering was probably when the whiners were in diapers any way, all the way back to 1970. <p>Foot pound doesn’t work for any engineer because 100 feet, and one pond doesnt tell the real story, but close enough is good enough these days?<p> Same goes for the English language, any thing approximate will tell the story because both parties only know half of what was taught in their class any way? <p>Strive to be better WILL, more literate, more accurate, and more educated, and don’t make up excuses lasting 492 words trying to explain why I got my degree in mechanical engineering and you didn’t.<p>And the watt still doesn’t include time. But it can be combined with time to form other words, like the Joule, and even other terms. <p>Thats like saying sugar is the cake, and that is just as absurd. <p>It takes sugar with other parts to make a cake, but sugar is still not a cake, its just one piece of the equation.<p>[ September 17, 2005: Message edited by: Chris Smith ]</p>

[rshayes]

There is an article on this subject, "Foot-pounds or pound-feet?", written by Richard L. Nailen, P.E. for a trade magazine (Electrical Apparatus, Sep 2004). It is available on line (http://www.findarticles.com/p/articles/ ... i_n9458018).<p>He makes it quite clear that he considers the use of "pound-feet" in the context of energy or power to be an indication of either ignorance or incompetence.<p>As noted before, several major handbooks (the CRC Handbook, the ITT Handbook, and Mark's) also use "foot-pounds" in the context of energy and power. The CRC Handbook was originally published in 1918, was still using this convention in 1967 (48th edition), and probably still does today. The ITT Handbook is a little newer, dating from 1942. Mark's dates from 1916.<p>The use of "pound-feet" does appear to be the convention in the context of torque. Unfortunately, torque was not the subject of the discussion.

[philba]

<blockquote><font size="1" face="Verdana, Helvetica, sans-serif">quote:</font><hr>Originally posted by stephen:

...
He makes it quite clear that he considers the use of "pound-feet" in the context of energy or power to be an indication of either ignorance or incompetence....
<hr></blockquote><p>we only get to pick one?

[Chris Smith]

Illiteracy is contagious. Torque is the factor from which all horse power is derived. With out the Torque there is no HP equation to derive any work from.<p>As Discussed already, you cant have 100 feet, and one pound, thus the foot pound statements are colloquial, inaccurate, and taught from one generation to the next with out question. This makes the subsequent generation Illiterate but they usually fight hard to keep that title? <p>Again, this doesn’t change the FACTS, its pound, foot for a logical and mechanical reason. <p>And James Watt was born in 1736, long before the CRC handbook started to misquote him and the physics equations.
********************
Foot-pounds or pound-feet?
Electrical Apparatus, Sep 2004 by Nailen, Richard L <p>
Use the terms interchangeably and people may question your technical competence<p>WHAT'S THE DIFFERENCE between foot-pounds and pound-feet? Both consist of the same units-a force and a distance-multiplied together. The physical quantities being expressed, however, are quite different. <p>"Yet applying the terms interchangeably is one of the most common misuses of technical language."<p>
The product of a force times a distance, as in "pound-feet" ("Newton-meters," for those metrically inclined), is torque. In electric machinery, we all know what torque means. It's the turning effort developed by a motor that causes a shaft to rotate. Based on the simple principle of the lever, torque is increased by either increasing the force itself (as by strengthening a magnetic field) or by increasing the distance-the lever arm -between the point efforce application and the axis of rotation. (Anyone using a wrench is familial with that concept.)
Any physics text or engineering handbook will make the principle clear and will define force times distancepounds times feet, or ounces times inches-as torque. To "say it right," remember the universal standard convention that the force comes first; the distance second.
So, is "foot-pounds" meaningless, then? Not at all. When the distance is named first, and the force second, the product represents something equally real but quite different: work.
Work measures the expenditure of energy. Thus, when a mass is lifted through a distance (against the force of gravity), or moved horizontally against a friction force, the amount of work done is measured by the product of the force involved and the distance moved. Increasing either one increases the amount of work.
How faxt that work is done is a measure of power (horsepower or watts). For example, if a 100-pound weight is lifted 5fi feet in one second, the work done is 550 foot-pounds, and the power involved is 550 foot-pounds per second-which equals one horsepower (746 watts).
Whether speaking of torque, then, as related to fastener tightening or motor capability, use the term putting force first (as in pound-feet). Putting the distance first (as in foot-pounds) is not equivalent; it represents something not at all related to torque.
And as we've said elsewhere and often, using technical terms incorrectly invites skepticism concerning your knowledge of technology.
By Richard L. Nailen, P.E., EA Engineering Editor
Copyright Barks Publications Sep 2004
Provided by ProQuest Information and Learning Company. All rights Reserved<p>[ September 18, 2005: Message edited by: Chris Smith ]</p>

[rshayes]

"Torque is the factor from which all horse power is derived. With out the Torque there is no HP equation to derive any work from."<p>I have never seen or heard of a horse with a driveshaft. Horses don't rotate and do not produce torque directly. A linear force can be converted to torque, but it is not necessary to do so to measure horsepower. There is indeed an equation for horsepower and it does not require torque.<p>
It is quite clear from the article that I cited and that Chris quoted that the conventional English unit for work is "foot-pounds". It also appears that this usage has been a convention for at least a hundred years (since both the CRC Handbook and Mark's use this convention). James Watt may or may not have used this convention or it may have been adopted at a later date.<p>Chris seems to feel that denying that a convention exists is a sign of a superior intellect that does not recognize mundane limits. Actually, it is a better indicator of ignorance and illiteracy.

[josmith]

I don't usually like to get into these peeing contests but I can't resist. I work for an old company who’s turn of the century product was a pump that used the linear force from a steam cylinder to move a piston which in turn pumped fluid. So tell me was it pumping without horsepower or was it producing horsepower without torque.<p>[ September 18, 2005: Message edited by: josmith ]</p>