The formula for computing ft/lbs energy from velocity and bullet weight? Lets do this in oz. grains, and feet. OK?

You can toss it out in metres/kilos if you want to show just how "international" you are. But ft/lbs is what I need.

Thanks

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PowderBurns Black Powder / Muzzle Loading Forum:

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A unit of work equal to the work done by a force of on point acting through a distance of one foot in the direction of the force. Energy in (ft lbs) therefore can be expressed as the product of the bullets mass and the ratio of the square of the velocity and its coefficent of reduction.

I think what you are looking for is the following:

e = w * (v^2/c)

where:
e = energy in ft lbs.
w = weight (and yes, you may use aviordupois as the unit of mass (instead of metric) as:
w = bullet weight (in grains) / 7000)
v = velocity squared (in fps)
c = coefficient of reduction is a constant set equal to: 64.32

Therefore if I have a bullet of 165 gr. (assuming sustained mass is retained fully upon exit at muzzle) proceeding to target @ 2,800 fps, the result set is 2,873 ft lbs.

Upon reading some site a few weeks ago however, I have learned that each of the models used to "predict" the energy in terms of knockdown power are all flawed. If I am able to find that site, I forward the URL address to you. Hope this helps you in the meantime.

Best regards,
Alex

The narrative is useful too. I'm no math whiz, and I've found that visualizing the functions helps to make sense of the forumula.

Who ever decided 7,000 gr. to a pound?

ft.lbs = weight of bullet (gr)x velocity (fps)x velocity (fps)/450240

450240 is a constant, haven't a clue what it does but it works every time.

[This message has been edited by 1894 (edited 06-21-2001).]

E = MC2/2. Thats mass x velocity square divided in half.

Energy in Joule = mass of bullet ( gramm ) x Velocity ( M / sec ) Square ( or x velocity again ) divided by 2000.

Why 2000: just for convenience: this way I may type in grams instead of kilograms ( 1000 ) times 2 because of the 2 in the equation.

Yeah, 7000 grs to the pound is really clever.

Have a good pocket calculator! H

Your number 450240 = 7000 X 64.32

7000 converts the grains to pounds.

A pound is a unit of weight, not mass. Units
of weight are based on the force of gravity on earth. That means that something weighing 100 pounds on earth will weigh less on the moon. 100 kilograms (a unit of mass) is the same on earth as it is on the moon. If I remember correctly, you must convert pounds to slugs (a unit of mass). The force of gravity is 32.16 ft/sec/sec. The constant 64.32 used above is 32.16 X 2.
Therefore pounds are converted to slugs by dividing by 32.16. As noted above, the whole equation is divided by 2 so using 64.32 is an easier way of accomplishing all of this.

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RC

[This message has been edited by rcasto (edited 06-21-2001).]

quote:

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Originally posted by aHunter:
<snip>
E = MC2/2. Thats mass x velocity square divided in half.
<snip>

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aHunter,

Just a heads up, I think what you are attempting to express is one-half the product of the mass of the body to the square of its velocity. Please note:

That would be expressed textually therefore as:

KE = 0.5 * m * (v^v) (or optionally
KE = (m * (v^v)) / 2

vice yours. Had you placed precedence around mv-squared, you would have completed the expression properly. Please check it out to verify my account. Is this correct?

One interesting note to share (not to digress) off the beat and path (a trivial pursuit or jeopardy question), but in the computer world, it is significantly costlier (in terms of processor speed and resources) to divide as opposed to multiply. It has nothing to do with the formula or problem, but it takes (on average) one-third the resources for calculators to multiply as opposed to divide. Hope that helps!

Best regards,
Alex

[This message has been edited by Alex Szabo (edited 06-21-2001).]

7000 gr. = l lb. works out nicely, but I wonder who came up with 437 1/2 gr. = 1 oz. ???

Never fails to amaze me which topics generate discussion and which one's just whither on the vine. That said, I'm gonna toss out some arcane trivia . . .

quote:

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Originally posted by PowderBurns:

7000 gr. = l lb. works out nicely, but I wonder who came up with 437 1/2 gr. = 1 oz. ???

Never fails to amaze me which topics generate discussion and which one's just whither on the vine. That said, I'm gonna toss out some arcane trivia . . .

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Yes Powderburns, The whole avoirdupois system escapes me. As I recall historically, the INCH came to be by the King declaring the width of his thumb as the fundamental linear unit. All I can say is that he must have had an enormous hand!

Hope you found the help here useful for updating your notes nevertheless.

Best regards,
Alex

[This message has been edited by Alex Szabo (edited 06-21-2001).]

KE = (1/2)mv*2

grains to pounds divide by 7000

weight to mass divide by the gravitational constant g = 32.16 ft/sec*2

gives ft.lbs. as the unit of KE.

That constant 450,240 is handy.

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Good huntin' and shootin',
RAB

On the subject of units for kinetic energy I have seen some confusion expressed (not here, but in magazines).

Mass (as correctely stated) is: force (pounds) divided by the acceleration due to gravity (32.174 feet per second per second), from Newton's First Law: F = ma. The standard units are slugs.

Velocity squared obviously is: ft^2/s^2

So why not slug-ft^2/s^2 ? I guess it just doesn't appeal. Its also hard to say without spitting.

The standard units are obtained by: Force (pounds) x Velocity (ft/s) x Velocity (ft/s) / Acceleration (ft/s^2), leaving the pounds intact and just cancelling the other terms.

Its a little screwy (like most things in English units) but its been the convention for a long time.

[This message has been edited by Harald (edited 06-22-2001).]

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Good huntin' and shootin',
RAB

Didn�t even had to upgrade the memory or a quicker processor for the division ;-)

H

KE = 1/2 mass x velocity^2.

V^V would be velocity to the velocity power, and 1/2mv*2 simplifies to just mv. 1/2mv**2 would be right.

As previously pointed out, you have to convert grains to pounds by dividing by 7039. Then you have to convert weight to mass by dividing by g, which is 32.174.

If you carefully do the equation, you'll be delighted to see that it yields 120 pounds of prime venison, two pheasant, and two terrific weeks in the field.

You would be amazed how many web sites give the conversion from grains to pounds as an even 7000. I seem to remember that was an approximation, but could not find any number other than 7000 even. I seem to remember 7040, but I'm unsure either way, now.

Thanks,
Don

[This message has been edited by Don G (edited 06-23-2001).]

You got me interested in this, so I had to dig out my old textbooks. The standard value of g that I quoted is for sea level at a latitude of 45 degrees. The text suggests that 32.2 ft/s^2 is sufficiently accurate for most analyses. The value varies with latitude, altitude, and also depending on whether you consider the rotation of the Earth or (properly) remove this effect. Most engineering analyses leave the rotation in because thats the practical measure, but a physicist would separate that effect since its an angular acceleration, not gravity.

I also misquoted Newton. F = ma is the SECOND law.

denton,
Mine was not a typo. "v*2" is an older method of showing "v squared". The asterix is not a multiplication sign in the format where it is elevated above the center line of the type line. If you have never been exposed to this convention, then I have educated you. However I am willing to accept the more modern form: "v^2".
I wish to know your source for 7039 grains per pound. The definition that I have always seen used is that 7000 grains is defined as one pound avoirdupois, exactly. It is a convention, as I understand it.

Well, that's it folks, 1001 Arabian posts!

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Good huntin' and shootin',
RAB

"v*2" is an older method of showing "v squared". The asterix is not a multiplication sign in the format where it is elevated above the center line of the type line. If you have never been exposed to this convention, then I have educated you. However I am willing to accept the more modern form: "v^2".

You are correct. Some users will be confused because of extensive use of BASIC and spreadsheet dialects. This is due to '*' = multiplication and '**' or '^' used to raise to a power.

Does Saeed have a tee shirt for those posting over a thousand? How about it Saeed?

[This message has been edited by Scott H (edited 06-24-2001).]

27.34 gr. = 1 dram

16 dram (437.44 gr.) = 1 oz.

1 lb would then equal 6999.04 gr.

-- This according to Webster's Unabridged, Twentieth Century Ed.

Of course all bets are off if you're heading East in a fast jet at 40,000 ft on a hot day in the summer, or trans-polar on a cold day in winter.

What I want to know is which has more foot pounds energy? (Notice I didn't write ft./lb.)

158 gr bullet @ 1450 fps, or 230 gr. bullet @ 980 fps ??? (Altitude, temperature, humidity, gravitational force, and earth's rotation being more or less constant . . . of course.)

Oh yeah, if you're traveling in a space craft at the speed of light and you turn on the headlamps . . . what happens?

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PowderBurns Black Powder / Muzzle Loading Forum:

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quote:

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Originally posted by fusilier:
Just for the curious, that 450,xxx figure is based on the constant of light speed, E=MC2. <snip>

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Fusilier, I am completely baffled by your 'input'. I was always under the impression that the speed of light was 186K ft/sec (299,792,458 meters/sec). Since we are all in accord that that value is exponentially squared, how did you arrive at your value since RCasto correctly identified the algebraic constant's value previously on the simplification that member '1894' presented?

Best regards,
Alex

Thanks for the the info on the *2 notation... never ran into that. The old Fortran (now I'm dating myself!) notation for exponent was '**', so I just assumed there was a missing asterisk, and that you had meant to indicate mulitplication.

quote:

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<snip>Oh yeah, if you're traveling in a space craft at the speed of light and you turn on the headlamps . . . what happens?

[/B]

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LOL PowderBurns! One may see their proximity as they materialize upon their arrival at the preeminent destination (which is to facilitate panned photography)!

Best regards,
Alex

Also, if we relly want to be precise then a method to calculate K.E. using a constant based on the universally accepted speed of light would be the way to go. I am very happy with using g = 32.174 ft/sec/sec for a handy constant of 450,436. Or let g = 32.16 and the handy constant is 450,240. Just be consistent.

I haven't had any physics or math courses for over 20 years and the only method of calculating kinetic energy I ever learned was the Newtonian KE = (1/2)MV^2.

Does anyone know the method of using the speed of light in a vacuum to compute KE? c = 186,282 miles/sec is all I know. That is something I memorized when I was 10 years old. I can't relate E = MC^2 to KE with out some tutoring or research. This is not fission or fusion, just internal combustion, so I am lost.

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Good huntin' and shootin',
RAB

quote:

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Originally posted by R. A. Berry:
O.K. guys if we really want to be precise then you must learn how to spell. Avoirdupois. Two of you above are mispelling it. One pound avoirdupois = 7000 grains, exactly.

Also, if we relly want to be precise <snip>

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It's "misspelling."

One pound avoirdupois is 699.04 gr. according to Webster's Unabridged.

Because you spelled it right in the first paragraph, I'm assuming the "relly" in the second paragraph is a typo.

Yeah, as a matter of fact I do teach English.

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PowderBurns Black Powder / Muzzle Loading Forum:

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BTW, nobody has taught me anything about the speed of light yet. I'm waiting for enlightenment.

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Good huntin' and shootin',
RAB

I'm gonna shoot the two different calibers into a couple sand bags. The one that makes the "most mess" is gonna get deemed the "baddest."

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PowderBurns Black Powder / Muzzle Loading Forum:

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The speed of light in a vacuum is always measured as the same, regardless of how fast the source of the beam is moving. It's counterintuitive, but has been very well demonstrated. If your spaceship is moving 99% of the speed of light, and you turn on the headlights, from your frame of reference, the light will still appear to you to be moving away from you at the same speed you would measure if you were standing still.

Of course the terms "moving" and "standing still" are also problematic. If you are in an enclosed box, moving through space, there is no experiment you can perform without outside references that will tell you whether you are moving or standing still. If the box is accelerating, there is no internal experiment you can perform to tell whether you are accelerating or in a gravitational field.

Aren't you sorry you asked?? :-))

quote:

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Originally posted by denton:
Speed of light:

Aren't you sorry you asked?? :-))

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OK, tell me this then . . .

If you dug a hole through the center of the earth and jumped in on one end would you come out on the other end or stop in the center?

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PowderBurns Black Powder / Muzzle Loading Forum:

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PowderBurn.....Shouldn't it be "badder" rather than "baddest"?

quote:

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Originally posted by DB Bill:
Correction to my earlier post.

PowderBurn.....Shouldn't it be "badder" rather than "baddest"?

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Perhaps, IF you subscribe to prescriptive grammars. Prescriptive grammars (contrasted to descriptive grammars) assert that there are rules to language which must be adhered to.

Prescriptive grammars derive from the grammars of Classical Latin and Greek which propose strict systems for the coherent and logical syntactical construction. Prescriptive grammars developed from the fields of logic and science. They were re-instated during the European Classical period wherein metaphysics were thought to adhere to strict systems of logic and order.
The shortcomings of such systems, when applied to vernacular English are that Latin and Greek are "organic" languages while English is a "synthetic" language.

Descriptive grammars are more anthropologically derived and seek to describe usage rather than "prescribe" rules. Without imposing structures, descriptive grammarians observe usage and propose social explanations for linguistic development.

Having never studied Classical Greek, let me use Classical Latin as an example: In Latin the statement "Tempus fugit." (Time flies.) can be rendered "Fugit tempus." with no change in meaning. Meaning in Latin is constructed through an organic grammar which inflects words according to their grammatical function.

English, by comparison, is syntactically governed. "Man bites dog." and "Dog bites man." have distinctly different meanings. The structure of each word is identical in both examples, but the syntax is different. Accordingly, the meaning is different.

The whole point of this aside is to illustrate that prescriptive grammars are necessary for organic languages, but they don't always work with synthetic languages.

Stanley Fish (Duke University), in Doing What Comes Naturally, posits that "interpretive communities" formulate systems (not rules) for communication. Use of such systems implies identity or "membership" in the group. Black English Vernacular (BEV -- some assert it doesn't exist) employs a grammatical system in English which varies from "standard English." (We're SURE that "standard English" doesn't exist! It's a Platonic ideal and hypothetical paradigm.)

One departure from "standard English" used in BEV is "ironic comparison" or "inverted description." It's theorized that American slaves developed this system to express feelings that were hostile to their overseers.

One obvious application of this usage is describing an object as "bad" when one means "good." E.g. "He plays one bad ax." This likely derived from the expression by slaves of such sentiments as: "My master is a kind, good man, and he takes special care of me."

We use irony in description all the time. I use it -- perhaps -- when I describe one caliber as "badder" than another when I mean that it is superior. Of course when I'm comparing destructive force, the use of "badder" to suggest "superior" is logical and not all that ironic.

But I use the superlative, "baddest." Prescriptive grammar holds that comparisons of two items requires use of comparative adjectives, not superlatives. A prescriptive grammar would isolate the statement from it's social context and consider only its syntactical content.

A descriptive grammarian would consider socio-linguistic context. We might argue that this comparison is the result of a hypothetical elimination of all other possible examples and therefore the "baddest" describes the outcome of numerous comparisons. Such comparisons might be stated: "Of the two 'baddest' options of all the available examples which one is the superlative?" The superlative would, of course be, "the baddest" and would be "the badder" of the two options.

BUT, the badder of the two options would be "the baddest" insofar as all other examples had been eliminated -- or insofar as the speaker wishes to imply that all other examples had been eliminated.

Finally, the speaker may simply wish to identify with a particular linguistic group ("interpretive community") and therefore chooses his words to "mark" his membership linguistically.

I'm using poetic license. "Badest" to imply that all options have been compared, but also to express a "near rhyme" allusion to "bad assed."

-- ya know what I mean, man?

OK, next question.

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PowderBurns Black Powder / Muzzle Loading Forum:

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Ironically edited for grammar and syntax.

[This message has been edited by PowderBurns (edited 06-27-2001).]

Just run that by me again???

Pete

quote:

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Originally posted by denton:
If you dug a hole directly through the center of the Earth (assuming for the moment that the Earth is not rotating), and jumped in, and if you could protect yourself from the ill effects of traveling through a few thousand miles of molten rock and metal, and if you evacuated all the air from the tube so there was no air resistance, you would pop out the other side in about 42 minutes. If you did not grab the rim, you'd get a chance to grab the rim at your starting point about 42 minutes after that.

---

OK, but would I be decelerating as I pass the mid point? Or would I continue to move at terminal velocity?

And you bring up another interesting issue: What's the difference between passing from pole to pole as compared to passing across the equator (other than the change in seasons )???

-- You know, this sort of discussion never occurs on the other forums. Mostly just talk about "crap hitting the propellers" . . . Nice contrast to consider the more esoteric and arcane.

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PowderBurns Black Powder / Muzzle Loading Forum:

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The Earth is not a true 'inertial reference frame' because it is rotating. For most day to day stuff, it is close enough. Technically, the tube experiment only works going from pole to pole. But since we're ignoring all kinds of practical difficulties, we might as well ignore the effects of rotation, too.

Now here's another one that has kind of a strange outcome: What if the Earth were hollow, and you got inside the shell. Would you feel any gravity? The answer is no. If you're inside a spherical shell, the attraction from shell neatly balances out, and you feel none.

Now you are on a role, anyone want to take a shot at "at what temperature does one cc of water weigh one gram". People lost interest and walked away from this one about 5 months ago.

Denton

What would your speed be as you passed through the Earths center.

Regards

Ray

quote:

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Originally posted by vssf:
Ok guys

Now you are on a role, anyone want to take a shot at "at what temperature does one cc of water weigh one gram". People lost interest and walked away from this one about 5 months ago.

Denton

What would your speed be as you passed through the Earths center.

Regards

Ray

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Zero degrees Centigrade . . . (just a guess).

You'd hit terminal velocity and maintain it to the center of the earth. Then you'd start deceleration, but I've no clue how that curve would look.

I tried the ft/lb formula E=w [v2 (squared)/c]

The results don't seem to agree with some online data from Marshall/Sanow

http://www.evanmarshall.com/towert/

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PowderBurns Black Powder / Muzzle Loading Forum:

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How fast would you be going as you passed through the center? That's a longer calculation than I want to make. Of course, if you don't evacuate the air, Powder is correct that you'd achieve terminal velocity, around 180-200 miles per hour, and stay at that speed until some time after you pass center. Of course, you would never reach the other side in that case.

In the airless case, the speed curve is just a simple sine wave, with the maximum and minimum corresponding to the times you arrive at the ends of the hole.