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Good huntin' and shootin',
RAB

I'm guessing 0 Centigrade because that's a "benchmark" for the metric system, as is 100 C.

A kilometer is based on some fractional distance from pole to pole.

Just to stir the pot, a nautical mile is one minute of latitude. 60 seconds equals a minute. 60 minutes equals a degree. 360 degrees in a circle.

Minutes and seconds derived from using a chronometer for navigation. If you know the Greenwich Mean Time (GMT), you can use a sextant and the position of the sun in the sky to calculate latitude/longitude -- but it's easier to use a GPS.

The last water temp/ volume /weight discussion went west after I suggested the correct temp was 15 Centigrade.

If you are using a sextant do you have to wait until local mid day i.e. when the sun is at its highest to establish your position?

Regards

Ray

quote:

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Originally posted by vssf:
PowderBurns

If you are using a sextant do you have to wait until local mid day i.e. when the sun is at its highest to establish your position?

Regards

Ray

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No, but you need to have a reasonably accurate watch. The more accurate the watch, the more accurate your position.

The sextant provides angle of the sun above the horizon -- which on the ocean is pretty constant. Time and angle provides latitude. Longitude is then calculated by using the angle of the sun, time of day, and compass bearing of the sun.

There's a relationship between minutes/seconds on a watch, minutes/seconds degree in angle, and the position of the sun in the sky. But I've no clue how that works, except that there are six degrees of angle in each minute on a watch.

When the sun goes down, you can shift to celestial navigation.

Navigational calculations are done with charts, a slide rule sort of instrument, or a navigational calculator.

If you know the height of an object (or width) you can use a range finding device called a kamal. This is essentially a calibrated stick (a ruler will work) held at a fixed distance and sighted on an object. Knowing the minutes of angle spanned by the object in the distance, and applying a coefficient for the increments on your kamal, you can estimate distance. On navigation charts, object heights -- even the height of bridges -- are provided so that the navigator can calculate distance.

This rangefinder system should work for shooting too.

My feeble mind recalled the approximation of about 40 F. I went and looked it up in "The Handbook of Chemistry and Physics." It's been a long time since I cracked that book.

While I was at it I looked up a formula for calculating the kinetic energy of an object based on the speed of light, c, as the constant. It is relativity based, and includes a term for the mass (m) at the given velocity and another term for the mass at rest ("m sub zero").

This formula is used for velocities that are a significant fraction of the speed of light. As the speed of the object approaches the speed of light, as we all know, its mass becomes infinite, and so does the kinetic energy of the object.

I am not going to try to reproduce this formula on this keyboard. I don't care about how somebody derived the 450,800 number (or thinks they did) for use as the handy constant for K.E. calculation. Adios amigos.

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Good huntin' and shootin',
RAB

I guess we rookies don't have a clue.
Barry

quote:

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Originally posted by 1604449:
You guys are a riot! From one simple equation to 4000 diversions. I've got tears in my eyes from laughing so hard. Where do you come up with this stuff!??

I guess we rookies don't have a clue.
Barry

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It's all in the ancient tradition of story-telling around the campfire. It's a tribal hunting ritual. In rural agricultural areas the "campfire" is the wood stove at the feed store. In suburbia it's the television. In cyberspace, it's the monitor and the forum.

-- Very important tradition . . .

Thank you I am enlightened.

Regards

Ray

"The sextant provides angle of the sun above the horizon -- which on the ocean is pretty constant."

I thought the sun would rise and fall the same as it does on land?

Regards

Ray

quote:

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Originally posted by vssf:
Powderburns

"The sextant provides angle of the sun above the horizon -- which on the ocean is pretty constant."

I thought the sun would rise and fall the same as it does on land?

Regards

Ray

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Well, the sun rises and sets, but the horizon is not constant on land. A mountain on the horizon will affect the perceived/computed angle of the sun in the sky.

A kayaker was paddling from San Fran. to Hawaii. After days/weeks out on the water, he was using his sextant to find his location. He found himself annoyed that the horizon was obscured by a landform . . . Then he realized that the landform was Hawaii.

He paddled onto the beach, secured his yak and fell into a sleep. He was awakening by some beach comber out for an early walk.

"Dude, you been paddling?"

"Yeah, I just paddled from San Fransico."

"Kewl . . ." and he walked on down the beach oblivious.

Maybe we need a forum for stories/tales.

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PowderBurns Black Powder / Muzzle Loading Forum:

www.hotboards.com/plus/plus.mirage?who=powderburns

The physics equation for kinetic energy is

e = � � m � v � v

when e = ft-lb, m = mass in pounds, and v = velocity

To convert this equation to a handier equation, one using weight instead of mass and grains instead of pounds, the gravitational constant (32.17398421) and the number of grains per pound (7,000) are easily and conveniently combined with the "2" in the "�" to form a single constant.

Without rounding anything off, the complete combined constant is 450,435.7789. But the gravitational constant is usually rounded-off � usually severely, often erroneously. With these different roundings-off, the combined constant that results is sometimes 450,240, sometimes 450,250, sometimes something else. I've even seen it as high as 450,800 somewhere.

Only the first three digits are really worth worrying about � 450,000 produces results as practical as any other, since the answer is going to be rounded-off, too.

Let's figure the energy of a 150-grain bullet at 3,000 ft/sec, using the approximations 450,000, 450,240, and 450,250, and the precise 450,436. The square of the velocity (9,000,000), the weight of the bullet (150 grains), and the product of the two (1,350,000,000) remain the same for each calculation.

1,350,000,000 � 450,000 = 3,000 ft-lb
1,350,000,000 � 450,240 = 2,998.400853 ft-lb
1,350,000,000 � 450,250 = 2,998.334259 ft-lb
1,350,000,000 � 450,436 = 2,997.096147 ft-lb

The last (2,997 ft-lb) is the most precise energy figure for a bullet that weighs exactly 150 grains, at the precise moment that it's going exactly 3,000 ft/sec. But given the fact that "150-grain" bullets vary in weight, and for only a moment does any of them ever go precisely 3,000 ft/sec, the figures of 2,997 to 3,000 ft/sec are all certainly close enough for me and the dogs I run with.

So the precise combined conversion constant is about 450,436.672, or (rounded-off) 450,437. And 450,000 is still close enough.

(Yeah, I know � BIG deal!)

It just dawned on me that we are limited to 3 significant digits by the bullet weight anyway. If one wanted to be a stickler he could use g = 32.174... and the handy constant of 450,437...but we would still end up having to round off the calculated energy figure to 3000 fps for that 150 grain bullet if we pay attention to significant digits, whether the constant chosen is 450,437 or 450,000.

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Good huntin' and shootin',
RAB

Now, as to Ken's question: I am not familiar with your twist nomenclature, Ken. If you could tell me what that means I will get to work on the calculation.

Anyway, if you have one of those 999 grain 220 Howell bullets handy, could you tell me how long it is?

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Good huntin' and shootin',
RAB