Talk:Degree of curvature

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Appearance[edit]

This article (page) opens poorly (is badly centered). Peter Horn 14:42, 10 October 2006 (UTC)

Degree of the Earths Cirvcumference: Generally taken as a proportion of 1/360 of the Earths equitorial circumference.

Eratosthenes is said to have measured and calculated the earths circumference supposedly the calculation is based on the assumption that the Earth is spherical and that the Sun is so far away that its rays can be taken as parallel.

Ertosthenes results, that make the degree 700 stadia, imply the circumference of the earth is 252,000 stadia and would be off by 1 part in 6 if the stadia he were using were Greek or Roman stadia of 185 m.

If his results were acurate his stadia would have measured 158.57 m. As it happens this works out to 302 Egyptian royal cubits.

According to Herodotus writing centuries before Eratosthenes in Book II of his histories the Egyptians had a very well documented standard of measure called the itrw which was 1/10 degree, related to both the Greek schoene and the Persian Parasang —Preceding unsigned comment added by 69.39.100.2 (talk) 22:53, 31 July 2008 (UTC)[reply]

Improvement of this article[edit]

This article is under the bad catagory of article it need serious improvements such as a good and unpartial definition. universal formula and some diagrams for the help to understing of an new person to this topic.Prymshbmg (talk) 14:52, 26 May 2015 (UTC)[reply]

This the theory which can be edited and refined to make a good article

A circular curve is often specified by its radius. A small circle is easily laid out by using the radius. In a mathematical sense, the curvature is the reciprocal of the radius, so that a smaller curvature implies a large radius. A curve of large radius, as for a railway, cannot be laid out by using the radius directly. We will see how the problem of laying out a curve of large radius is solved. In American railway practice, the radius is not normally used for specifying a curve. Instead, a number called the degree of curvature is used. This is indeed a curvature, since a larger value means a smaller radius. The reason for this choice is to facilitate the computations necessary to lay out a curve with surveying instruments, a transit and a 100-ft engineer's tape. It is more convenient to choose round values of the degree of curvature, rather than round values for the radius, for then the transit settings can often be calculated mentally. A curve begins at the P.C., or point of curvature, and extends to the P.T., or point of tangency. The important quantities in a circular curve are illustrated above.

The degree of curvature is customarily defined in the United States as the central angle D subtended by a chord of 100 feet. The reason for the choice of the chord rather than the actual length of circumference is that the chord can be measured easily and directly simply by stretching the tape between its ends. A railway is laid out in lengths called stations of one tape length, or 100 feet. This continues through curves, so that the length is always the length of a series of straight lines that can be directly measured. The difference between this length, and the actual length following the curves, is inconsequential, while the use of the polygonal length simplifies the calculations and measurements greatly.

The relation between the central angle d and the length c of a chord is simply R sin(d/2) = c/2, or R = c/(2 sin d/2). When c = 100, this becomes R = 50/sin D/2, where D is the degree of curvature. Since sin D/2 is approximately D/2, when D is expressed in radians, we have approximately that R = 5729.65/D, or R = 5730/D. Accurate values of R should be calculated using the sine. For example, a 2° curve has R = 2864.93 (accurate), while 5730/D = 2865 ft.

If some other value and length unit are chosen, simply replace 100 by the new value. In the metric system, 20 meters is generally used as the station interval instead of 100 ft, though stations are numbered as multiples of 10 m, and these equations are modified accordingly. With a 20 m chord, R = 1146/D m,or about 3760/D ft. Of course, a given curve has different degrees of curvature in the two systems. There are several methods of defining degree of curvature for metric curves. D may be the central angle for a chord of 10 m instead of 20 m.

The deflection from the tangent for a chord of length c is half the central angle, or δ = d/2. This is a general rule, so additional 100 ft chords just increase the deflection angle by D/2. Therefore, it is very easy to find the deflection angles if a round value is chosen for D, and usually easy to set them off on the instrument. For example, if a curve begins at station 20+34.0 and ends at station 28+77.3, the first subchord is 100 - 34.0 = 66.0 ft to station 21, then 7 100 ft chords, and finally a subchord of 77.3 ft. The deflection angle from the P.C. to the P.T. for a 2° curve is 0.660 + 7 x 1.0 + 0.773 = 8.433 °, or 8° 26'. I have used the approximate relation δ = (c/100)(D/2) to find the deflection angles for the subchords.

The long chord C from P.C. to P.T. is a valuable check, easily determined with modern distance-measuring equipment. It is C = 2R sin (I/2), where I the total central angle. For the example, C = 2(2864.93)sin(8.433) = 840.32 ft. The length of the curve, by stations, is 843.30 ft. This figure can be checked by actual measurements in the field. The actual arc length of the curve is (2864.93)(0.29437) = 843.34 ft. Note that this is the arc length on the centre line; for the rails, use R ± g/2, where g = 4.7083 ft = 56.5 in = 1435 mm for standard gauge.

Before electronic calculators, small-angle approximations and tables of logarithms were used to carry out the computations for curves. Now, things are much easier, and I write the equations in a form suitable for scientific pocket calculators, instead of using the traditional forms that use tabular values and approximations.

A 1° curve has a radius of 5729.65 feet. Curves of 1° or 2° are found on high-speed lines. A 6° curve, about the sharpest that would be generally found on a main line, has a radius of 955.37 feet. On early American railroads, some curves were as sharp as 400 ft radius, or 14.4°. Street railways have even sharper curves. The sharpest curve that can be negotiated by normal diesel locomotives is not less than 250 ft radius, or 23°. It is not difficult to apply spirals, in which the change of curvature is proportional to distance, to the ends of a circular curve. Circular curves are a good first approximation to an alignment. — Preceding unsigned comment added by Prymshbmg (talk • contribs) 14:55, 26 May 2015 (UTC)[reply]

If I recall correctly, the chord used in Mexico is 20 m or 65.62 ft. Peter Horn User talk 15:08, 1 June 2015 (UTC)[reply]

Another formula for chord length[edit]

Degree of curvature#Formula for Chord length

C

is chord length

R

is radius of curvature

Dc

is degree of curvature, chord definition

Using the trig formula for any triangle: a/sinA=b/sinB=c/sinC, then R/[sin(180°-DC)/2]=C/sinDC. R=Cx[sin(180°-DC)/2]/sinDc. At this point I need assistance to put this in the $sample$ format. This is an isosceles triangle with two sides being R and two angles being (180°-Dc)/2. Peter Horn User talk 16:05, 1 June 2015 (UTC)[reply]

@Peter Horn: Please include a reference along with its theory and diagram.Prymshbmg (talk) 18:10, 1 June 2015 (UTC)[reply]

@Peter Horn: Also it will get converted into the given formula as proof given below.

{\frac {R}{\sin({\frac {180-D_{C}}{2}})}}={\frac {C}{\sin(D_{C})}}\Rightarrow {\frac {R}{\sin(90-{\frac {D_{C}}{2}})}}={\frac {C}{\sin(D_{C})}}\Rightarrow {\frac {R}{\cos({\frac {D_{C}}{2}})}}={\frac {C}{\sin(D_{C})}}

\Rightarrow {\frac {\sin(D_{C})}{\cos({\frac {D_{C}}{2}})}}={\frac {C}{R}}\Rightarrow {\frac {2*\sin({\frac {D_{C}}{2}})*\cos({\frac {D_{C}}{2}})}{\cos({\frac {D_{C}}{2}})}}={\frac {C}{R}}\Rightarrow 2*\sin({\frac {D_{C}}{2}})={\frac {C}{R}}

\Rightarrow R={\frac {\frac {C}{2}}{\sin({\frac {D_{C}}{2}})}}

This is same formula as given in the article.Prymshbmg (talk) 18:41, 1 June 2015 (UTC)[reply]

The diagram requested is to be found in the isosceles triangle article. What you get here is an isosceles triangle of which one side, the chord, equals 100 ft (30.48 m) and the other two (2) sides equal R. No more, no less. Any old fashioned trigonometry textbook gives the theorem a/sinA=b/sinB=c/sinC, that is the sides divided by the sine of the opposite angle. No need to get the cosine involved. Peter Horn User talk 23:28, 1 June 2015 (UTC)[reply]

@Prymshbmg:

Eureka, Triangle#Sine, cosine and tangent rules

Peter Horn User talk 23:52, 1 June 2015 (UTC)[reply]

I found a better diagram. Using the law of sines, a curve of 21° (ALCO RS-3) gives R/[sin(180°-21°)/2]=100ft/sin 21°. Solving for R, using my trusted TI-36x SOLAR, i get a radius of 274.3702133 ft. The calculator does no rounding off. Peter Horn User talk 01:15, 2 June 2015 (UTC)[reply]

I have added another image to this discussion. Peter Horn User talk 13:55, 2 June 2015 (UTC)[reply]

To sum it up, R/sin79.5° = Chord/sin21° Therefore R= (Cord x sin 79.5°)/sin21° . The sample calculation in the article and above is unnecessarily convolute. way too convolute. Peter Horn User talk 14:09, 2 June 2015 (UTC)[reply]

@Peter Horn: Please see the above proof and close the discussion. As you can see that formula in the article is for generalised use and can be used any where, with some constants and doesn't need revision. Thanks for the discussion and Happy Editing.Prymshbmg (talk) 16:47, 2 June 2015 (UTC)[reply]

I think a few elements of the long "theory which ... can be added" may be useful, but would not favor inserting that entire writeup. The existing formulas are good and I don't think we need a derivation or further numerical examples. None of the figures posted here show the actual relationship of the angle and curve, and finding one might be a useful addition.

I wish there was a good place to put a summary of all the equations for computing and laying out a circular curve, but that seems outside the scope this article. BillHart93 (talk) 20:49, 2 June 2015 (UTC)[reply]

@Prymshbmg: With all due respect, effectively to tell me to shut up is rather pompous on thine part and is almost tantamount to a personal attack and at least uncivil. In English the expressions a second or an alternativ are synonymous with another. This means that the second method/formula in no wise invalidates the first method/formula and therefore there was absolutely no need for proofing/validating the original formula. In effect there are two (2) valid formulae. kind regards. @BillHart93: Peter Horn User talk 14:59, 3 September 2015 (UTC)[reply]

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This does not work[edit]

Degree of curvature#Formula from chord length $r={\frac {C}{2\sin \left({\frac {D_{\text{C}}}{2}}\right)}}$ does not give a correct answer but a radius that is half of what it should be.
$r={\frac {C}{\sin \left({\frac {D_{\text{C}}}{2}}\right)}}$ gives the correct answer for a 50 ft (15.24 m) chord. Peter Horn User talk 01:00, 7 August 2021 (UTC)[reply]

@Peter Horn: I looked at the trig, and I think you're mistaken. You have to divide the chord length by two to create the right triangle from the isosceles triangle. But it's been a lot of years since I did many sines and cosines, so I'm open to the notion that I goofed somewhere. Indefatigable (talk) 16:34, 7 August 2021 (UTC)[reply]

@Indefatigable: Your memory of trig is correct, so the formula should be adjusted accordingly. However,the original formula kept giving me a radius half of what it should be. Peter Horn User talk 18:51, 7 August 2021 (UTC)[reply]

@Indefatigable: My T(exas) I(instruments) 36X SOLAR appears to have gone bonkers. Peter Horn User talk 19:43, 7 August 2021 (UTC)[reply]

@Indefatigable: The original formula gives me a radius of 239.17 feet for a 12 degree angle, it should be double that. Peter Horn User talk 20:25, 7 August 2021 (UTC)[reply]

When C = 100 ft and D_C = 12°, the formula with 2 in the denominator yields 478.3 ft. I'm wondering why you're using a 50 ft chord - isn't that unusual? I'm not a surveyor, but aren't 100 ft chords the common practice? And where are you getting the "correct" answer from? A book of surveyor's tables? Based on 100 ft chords? Indefatigable (talk) 21:18, 7 August 2021 (UTC)[reply]

Of course, the chord is 100 ft. Why did I get hung up on 50 ft chord? Peter Horn User talk 23:08, 21 August 2021 (UTC)[reply]

Curvature and Curved screens[edit]

Should we add a link to the Curved screens article from here? 75.142.198.200 (talk) 04:17, 27 November 2021 (UTC)[reply]

That article is irrelevant. Peter Horn User talk 04:46, 27 November 2021 (UTC)[reply]