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    I work in the rail industry and I am trying to prove a rail has been curved correctly to a Radius. On the picture the rail would be the arc for which i have a length. it would be located between the 2 points of the chord, effectively creating a segment.
    I am also given the circle radius the rail is to be curved to. The chord would be a string line used to measure to. this is how we check the curve is correct.
    The arc length is split into 4 equal parts and marked giving me 3 measurement to check at the 1/4 1/2 and 3/4 of the length of the arc. measured from each of the 3 points marked along the arc or in my case a rail, square or at a right angle to the chord or in my case string line.
    I am given the arc length and radius how would I calculate the distance between the arc and chord (running at 90 degrees to the chord) at the 1/4 1/2 and 3/4 points of the arc measurement.

    I have tried to explain this as best I can and any help would be appreciated. Thanks in advance
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    do you mean the 3 lengths at the top, on the inside?...
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  1. File Type: pdf sunday.pdf (1.2 KB, 59 views)
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    (Original post by Hasufel)
    do you mean the 3 lengths at the top, on the inside?...
    the three positions at the top on the arc are spot on but where you have them going to the centre i want the distance from the point on the arc vertically straight down to the chord line (highest horizontal line on your picture). the distance is measured at a right angle to the chord in the practical application.
    hope this helps you solve my problem Thanks
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    I know that, if you have half the chord length you can get the middle bit, called the "Sagitta":

    http://liutaiomottola.com/formulae/sag.htm
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    This could help me but I'm not given the chord length (l). I'm given the arc length, if that helps?
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    I can`t yet find a formula for the sagitta with only the arc and raduis, then the "drop-down line" - here we`ll call is "D" is :

    \displaystyle D = r+ \frac{r}{2} \sec(\theta) \left( \cos \left(\frac{3 \theta}{4} \right)+\sin \left(\frac{5 \theta}{4} \right)-2 \sin(\theta) \right)

    (i am nearly bald after working this out!)

    (here, you don`t even need the arc!)

    EDIT: balls! - i`m just going to double check...

    (doesn`t work for angles of (3n+1)Pi/2,n=0,1,2...) so there is obviously another way of doing this - knackered if i can find it!

    Bed time!
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    I honestly think my formula (above) has a mistake in it, but i`ll keep working at it - something to do with the angles involved.

    I`ve found what the sagitta is:

    \displaystyle S= r \left(1- \cos \left(\frac{\theta}{2}\right) \right)

    Beg pardon...=radius - apothem

    can get theta from arc, since Arc= r times theta (in radians of course)

    EDITED
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    http://burymathstutor.co.uk/railarc.html

    ?
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    EUREKA!!!

    (my first post on distance was too complicated, and i suspected a mistake - the answer is soooo simple!)

    you have the raduis and the arc, right, so you can get the angle theta (radians):

    from a=r \theta

    1) imagine the triangle formed by the 2 "outer" radii and their chord - the height of this triangle is called the apothem of the circle for the given angle \theta.

    the solution is just the difference in heights - or apothems - of this triangle, and the one formed by the 2 "inner" radii.

    the 1st triangle (we`re looking at half of it) has height:

    \displaystyle r \cos \left(\frac{\theta}{2} \right)

    since the angle is now halved.

    the 2nd has height (or apothem - again looking at half of the INNER triangle):

    \displaystyle r \cos \left(\frac{\theta}{4} \right)

    Therefore, the solution you want is the difference in apothems:

    \displaystyle D_{istance}= r \left( \cos \left(\frac{\theta}{4} \right)- \cos \left(\frac{\theta}{2} \right)\right)

    and since from the formula for arc length,: a=r \theta

    this gets:

    \displaystyle D_{istance}= r \left( \cos \left(\frac{a}{4r} \right)- \cos \left(\frac{a}{2r} \right)\right)

    THIS has been great! - now i can finally sleep without something bugging me!
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    (Original post by Hasufel)
    EUREKA!!!

    ...


    THIS has been great! - now i can finally sleep without something bugging me!
    I wrote a little program for the problem. An interesting application of topics from C1.



    I may have understood the OP's requirements but my program doesn't do what your formula does. Are we solving different problems?

    Edit: I have solved the wrong problem. I will modify my solution.
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    Fixed. We now agree.

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    [QUOTE=BuryMathsTutor;53727535]Fixed. We now agree.=QUOTE]

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    Absolutely Fantastic. Thanks you Hasufel and BuryMathsTutor, The image in post 11 is exactly what I'm looking for and the program makes it so easy to apply the maths quickly and simply in my work environment.

    Using the program I can now find a measurement to check against at any point along the rail (arc) quickly and easily, which will help me to build a better quality product.

    I will be trying this when I return to work on Monday.

    One question. Is there any way I can download the railarc.html web page to use off line?

    Thanks again for your help.
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    (Original post by TTBoy)
    Absolutely Fantastic. Thanks you Hasufel and BuryMathsTutor, The image in post 11 is exactly what I'm looking for and the program makes it so easy to apply the maths quickly and simply in my work environment.

    Using the program I can now find a measurement to check against at any point along the rail (arc) quickly and easily, which will help me to build a better quality product.

    I will be trying this when I return to work on Monday.

    One question. Is there any way I can download the railarc.html web page to use off line?

    Thanks again for your help.
    No problem. You can save a complete webpage easily enough. Try pressing ctrl and S. You can even save webpages on some of the better mobiles. May I ask what your job is? My students sometimes ask questions like, "What could this be used for?" I have some answers for these questions but it would be good to have another.
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    (Original post by BuryMathsTutor)
    No problem. You can save a complete webpage easily enough. Try pressing ctrl and S. You can even save webpages on some of the better mobiles. May I ask what your job is? My students sometimes ask questions like, "What could this be used for?" I have some answers for these questions but it would be good to have another.
    Thanks again. I saved the complete web page, it wasn't until I'd had a couple of hours sleep I realised I didn't have to be online to use it in a browser lol.

    I work for a company that supplies specialist track work for the rail industry. The majority of our work is for Network Rail here in the UK. We also supply London Underground, MTRC in Hong Kong, Amtrak in the USA, SMRT Corporation in Singapore and Eurotunnel.
    Our facility supplies renewals and maintenance track work, sets of switches and crossings to either be supplied straight to site or to our renewals facility that build complete turnouts, crossover and junctions as they would be in the rail network which are then inspected by our customers before being transported for installation into the rail network.
    I work mainly on the final switch assembly after all the component parts have been manufactured on site. Each switch has to be build to specific dimensions and tolerances as set out by the customer.
    Product quality is critical to ensure the safety of product once installed.

    It's not a particularly complicated process, the tolerance on the distances the program calculates is +/- 1mm and the rails are very flexible before they are fixed to timber sleeper or concrete bearers.

    Thanks
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    (Original post by TTBoy)
    Thanks again. I saved the complete web page, it wasn't until I'd had a couple of hours sleep I realised I didn't have to be online to use it in a browser lol.

    I work for a company that supplies specialist track work for the rail industry. The majority of our work is for Network Rail here in the UK. We also supply London Underground, MTRC in Hong Kong, Amtrak in the USA, SMRT Corporation in Singapore and Eurotunnel.
    Our facility supplies renewals and maintenance track work, sets of switches and crossings to either be supplied straight to site or to our renewals facility that build complete turnouts, crossover and junctions as they would be in the rail network which are then inspected by our customers before being transported for installation into the rail network.
    I work mainly on the final switch assembly after all the component parts have been manufactured on site. Each switch has to be build to specific dimensions and tolerances as set out by the customer.
    Product quality is critical to ensure the safety of product once installed.

    It's not a particularly complicated process, the tolerance on the distances the program calculates is +/- 1mm and the rails are very flexible before they are fixed to timber sleeper or concrete bearers.

    Thanks
    Thanks.
 
 
 
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