Evan and Company 
I've been thinking about ArcChord Separation question, and have come to believe that the 0.00094+ inch separation over an 80 foot length overstates the true value by a factor of two.
The "formula" is ArcChord Separation = Radius x [1  Cosine (Subtended Angle / 2)], which would be 6378 kilometers x [1  Cosine(0.000219 degree / 2)] = 6378 kilometers x [1  Cosine(0.00011 deg)].
An alternative method of calculation would be Separation = (Chord Length / 2) x Tangent (Subtended Angle / 4) = 40 feet x Tangent (0.000219 degree / 4).
The first calculation is somewhat treacherous on to perform on a pocket calculator, because of limited accuracy in the computation of Cosines of very small angles. Done in Excel, though, both algorithms provide a calculated result of a wee bit less than 0.0005 inches, as does a very inexpensive Casio calculator only if using the second calculation.
Why? The Sines and Tangents of Small Angles are very nearly linear functions of the Small Angle magnitude, but not so with the Cosines of Small Angles. The Small Angle Approximation for Cosines is [1  Cosine(Small Angle)] is approximately equal to 4 x [1  Cosine(Small Angle / 2)].
John
Announcement
Collapse
No announcement yet.
About Those Carpenters' Levels
Collapse
X

Geodesy
What we are talking about here is geodesy, which is often confused with but is related to geodetics and used in surveying and GPS  amongst others.
Most Land survey work is in relatively small "parcels" in which the difference between the distance between 2 points measured over the chord (straight line) and curve/arc (of the earth surface) is so small that any differences distance and in elevation (height)  both real and relative  are so small as to be of no consequence.
Thus most land survey work is "Plane" survey which assumes a true horizontal plane.
When surveying over longer distances and/or larger areas the differences do become significant and so the survey becomes a geodetic work where the survey is conducted on and between points on a sphere. Spherical geometry is used instead of plane geometry. Gravity becomes an issue as well.
The curvature of the earth is most noticeable at the horizon.
Geodetic survey work requires and uses instruments and measurements of such precision that we can only contemplate them.
Astronomy is the same  sort of  mostly. Evan and others have demonstrated that as well over time in several previous threads.
So Forrest was quite correct and perhaps inadvertently demonstrated the relevance of geodetics.
This is what Forrest ran into on his machine bed. To do it once was extraordinary  but to repeat it was fabulous!!
That was damn fine leveling (surveying??) Forrest!!!
Leave a comment:

Originally posted by Optics CurmudgeonA purely trigonometric analysis gives a value of 3.129E7 inches drop over 1 foot
"Because of the curvature of the Earth, a level displays an error of approximately .00003" per foot. With this in mind, a surface that is 10' long when scraped or adjusted to indicate dead level along its entire length is actually .00015" low on each end. This is a minor problem with a surface this small but when setting up a machine 60' long, this curve begins to be an appreciable problem. With a machine of this length, the error on each end would be .0009" or nearly one thousandth of an inch. If the machine were required to hold part tolerances of less than .0005" this would be a problem."
By the way, that quote is out of the chapter on Machine Alignment, which is outstanding.
Leave a comment:

"BTW, metric is sooo much easier for this sort of calculation." Maybe so but how do you convert back to cubits and quintals and other units what I was fetched up to unnerstand.?
Leave a comment:

6378000 meters radius
40074155.2056 meters circumference
111317.09779 meters per degree
80ft= ~24.384 meters
.000219 degrees per 24.834 meters
cos (.000219)x radius = 6377999.9999534094878915819570174
Difference is .00004659051210841804298257045375234 meters
The rule of small angles say we can call it a triangle so divide by two to get .00002395256 meters or .02395256 millimeters
.02395256 millimeters = 0.000943014173 inches
Hey, wadda ya know? That's close enough to be called agreement.
BTW, metric is sooo much easier for this sort of calculation.Last edited by Evan; 04162008, 07:10 PM.
Leave a comment:

Fearing the worst, I've gone back and redone my original numbers. Originally I was using a spreadsheet I set up for a spherometer, which uses a "less rigorous" calculation, expecting the ratio of R to r to be relatively large compared to what we are discussing. The result is an increasing error as we get to very small angles. A purely trigonometric analysis gives a value of 3.129E7 inches drop over 1 foot, and for Forrest, a value of .000935 high in the center of an 80 foot span. That's really good, using levels, regardless of their rated accuracy. Of course, I could still be fooling myself somehow, and invite independent confirmation.
Joe
Leave a comment:

Originally posted by Forrest AddyIn 1996 I moved a big floor mill across a customer's shop to a foundation which had been prepared for it. It was quite an evolution but in the end the 80 ft long runway was leveled as close as my helper and I could get it using a couple of Taft Pierce 6 arc second levels using reveral technique.
The laser showed my work to be slightly high in the center. The tech and I were interested in the reason for the difference but cause both laser and level showed consistant and repeatable readings and there were no significant heat sources to interfer with our readings. We checked and rechecked. It was the laser guy who suggested we discovered the difference was the Earth's chordal height in 80 ft
Leave a comment:

I guess its time to recount an experience I had.
In 1996 I moved a big floor mill across a customer's shop to a foundation which had been prepared for it. It was quite an evolution but in the end the 80 ft long runway was leveled as close as my helper and I could get it using a couple of Taft Pierce 6 arc second levels using reveral technique. The bedways were set on grouted wedge levelers a system far more rigid than leveling screws.
Installation and alighment proceeded from there. In the end the customer contracted another outfit with a laser system to verify what i had done. No problem from me. I was confident in my work and interested in a comparison of the two systems. The laser showed my work to be very slightly high in the center. The tech and I were interested in the reason for the difference. Both laser and level showed consistant and repeatable readings. There were no significant heat sources to interfer with our readings. Our readings showed consistancy not enough to actually quantify the chordal height. We checked and rechecked. It was the laser guy who suggested the difference was the Earth's chordal height in 80 ft: we had verified the Earth really was round  or local gravitation had a lump in it.
Makes sense. Levels refer to gravity and define a spherical surface perpendicular to the local force of gravity. Lasers produce a beam of coherent light which defines a line having practically zero curvature in the Earth's gravitational field. If both apparatus are sufficiently sensitive, 80 feet will be enough reckoning distance for the difference between the two systems to become apparent. An arc second along the Earth's surface by the way is about 101 feet  1/60 of a nautical mile. So what is the chordal height of a 4000 mile radius in 80 feet? Nah! The height in that distance is vanishingly small but why did we get the consistant readings? Another machine shop mystery.Last edited by Forrest Addy; 04162008, 03:00 PM.
Leave a comment:

.0015 inches per foot? Better check that math, that implies that the Earth has a radius of 1000 feet. Using an average radius of 3955 statute miles, the real value is more like 8.62E7 inches per foot, or 0.862 microinches.
Joe
Leave a comment:

Hmmm. Just how, I wonder? If the level is used to detect the planarity of the ways in the same manner as I did then a change in the gravity normal caused by the curvature will not affect the reading in measurable proportion to the change. Only if the level is turned 90 degrees to measure along the length of the bed will a change be seen, and that assumes the lathe bed is tangent to the gravity normal. Measurements perpendicular to the ways will read the same even if the lathe were long enough to encircle an ideal smooth Earth.
The only way that a deviation due to the curvature of the Earth would be noted in the perpendicular orientation would be if the bedways deviated from straight in the horizontal plane. The deviation would then be very hard to detect because it would be equal to the sine of the angle that the ways deviate from straight times the total deviation expected from the curvature. This would be many orders of magnitude below the detection ability of the best level.
Deviations from true planarity in the longitudinal direction are of far less importance since they will only affect the tool height in respect of the work. Even that assumes that the lathe extends at a tangent to the gravity normal and even then the affect on a level in the perpendicular orientation is minute.
In other words, the change in gravity normal with length will not affect the proper use of a level in determining the planarity of the lathe bedways perpendicular to the turning axis no matter how large the lathe. The effect over distance on planarity in the longitudinal direction will be so small to be of no consequence in nearly all cases. The 50 foot lathes that I routinely observed in operation at Cariboo Pulp and Paper would have a maximum tool height error of only about .002" at the tailstock, an insignificant amount when turning a 6 foot diameter roller 40 feet long. That assumes the roll stays perfectly straight in respect of the lathe.Last edited by Evan; 04162008, 07:14 PM.
Leave a comment:

Originally posted by Peter NeillThe curvature being approximately 0.0015" every 12".
Mike is a professional machine tool rebuilder, and he was pointing out that when you align and rescrape a large bed mill, grinder, or lathe, that the ways are easily long enough that the .0015" per foot is enough to induce a pretty significant error.
Leave a comment:

This reminds me of an argument I use to illustrate the fallacy of astrology. People think that the gravitational influence of Jupiter is important to how their life will turn out when in fact the influence of the milk delivery truck that was parked in the driveway was much greater.
Leave a comment:

As long as a fresh bourbon and coke don't spill, or tump over, when you set'er on the ways, I say "Gitter Done"
Leave a comment:

Originally posted by EvanMore important in John's case is the effect of what are called Masscons, or Mass Concentrations. They use flying gravitometers to find such things as large quantities of buried iron.
.
Leave a comment:

The curvature of the Earth is much easier to see than most people think. There is a beach on Murtle Lake here that is perfect for observing the curvature. When sitting on the sand you cannot see the white sand beaches on the opposite shore about 5 kilometers distant. Stand up and they are clearly visible.
More important in John's case is the effect of what are called Masscons, or Mass Concentrations. They use flying gravitometers to find such things as large quantities of buried iron.
Leave a comment:
Leave a comment: