Mark Mayer, post: 443625, member: 424 wrote: It's better if the customers don't see how the sausage is made.
We work such a tight schedule in construction that it is imperative to be in constant communication with not only field superintendents, but also the office personnel associated with each project on a daily basis.
People who blame others because of their own personal inabilities are total Jackasses.
I suppose business methodology is different depending on the circumstances. 😎
John, post: 443652, member: 791 wrote: Though I drew the arcs to the specs, neither arc length agreed with the drawing provided. I had the chord length and radius correct, but one arc length was off about half an inch, the other further (don't remember how much off the top of my head).
All the numbers should agree between my drawing and the PDF, but I can not make them all match. Should someone here draw them in cad, perhaps then we can attempt to find out what is really going on. The numbers generally point in the correct direction, but, well, if I can't get all numbers to match with the info given, something is wrong.
Missing by a half inch is likely a rounding error in the curve data and pretty much meaningless if the delta is rounded to the nearest second. Pretty standard on a lot of older plats to just get arc length and radius and sometimes not even the radius. If you have the incoming and outgoing tangent bearings along with arc length, everything else is easily derived with simple trig. Funny thing about this, in all my years I have yet to recover any survey control in the field that matches 'exactly' the curve data provided. If I match within a half an inch on the ground, I dance a little jig.
Williwaw, post: 443664, member: 7066 wrote: Missing by a half inch is likely a rounding error in the curve data and pretty much meaningless if the delta is rounded to the nearest second. Pretty standard on a lot of older plats to just get arc length and radius and sometimes not even the radius. If you have the incoming and outgoing tangent bearings along with arc length, everything else is easily derived with simple trig. Funny thing about this, in all my years I have yet to recover any survey control in the field that matches 'exactly' the curve data provided. If I match within a half an inch on the ground, I dance a little jig.
Fair enough, had not considered rounding errors in this process. And when I am provided minimal info to start with (Only radius and arc length, nothing else at all), I wanted something else to run with as well. When I got the chord distance, the arcs didn't agree. And the sales person is convinced I have a magic button in CAD to draw whatever I tell it to draw. (but he won't tell me where that magic button is since he has never used cad a day in his life). I had other work to do and didn't consider other ways to get things done besides for ask for a simple piece of the puzzle since the customer could not be bothered to supply the cad drawing itself (and the sales person could not be bothered to ask)..... which would have saved a week or more of messing around with getting things done.
Curvefromendofobject. All one word.
I worked both of the parts that are in your other thread, the numbers are accurate to within a tolerance of 1/64". Bear in mind that given any 2 pieces of curve data you should always be able to calculate all of the remaining data using standard, relatively simple curve formulas. The fun exercise is to sit down with a basic sketch of curve elements and derive the formulas on your own. The misleading thing about the sketches that you were provided is that the curve is nearly 180 degrees not the more or less 45 degrees that they show in the sketch.
thebionicman, post: 443672, member: 8136 wrote: Curvefromendofobject. All one word.
Unknown command for Autocad LT. Using LT does present some limitations, but this company truly can not make use of Civil 3d and would rather not incur the extra cost of something we have little use for.
Randy Rain, post: 443678, member: 35 wrote: I worked both of the parts that are in your other thread, the numbers are accurate to within a tolerance of 1/64". Bear in mind that given any 2 pieces of curve data you should always be able to calculate all of the remaining data using standard, relatively simple curve formulas. The fun exercise is to sit down with a basic sketch of curve elements and derive the formulas on your own. The misleading thing about the sketches that you were provided is that the curve is nearly 180 degrees not the more or less 45 degrees that they show in the sketch.
Interesting. Now I'm wondering why I could not get closer than I did using Autocad and the commands available to me. Gonna have to look at that some more.
for sure DON'T need C3D for something like that! Just draw the circles. One for the radius of the part, one for the chord distance and check against the Length.
Since this IS NOT a surveying issue, but a machining one, just How accurate is the setup?? Your distance to a 16th of an inch is realistic for building?
I believe the machine cutting tolerance is somewhat greater than 1/8th of an inch. I have finally learned to basically ignore 1/16th errors.
You are quite right, with the chord distance and radius provided, drawing such things is Easy. Initially, I did not have the chord distance. Just the radius and arc distance.
AHhhhh ! that's right! Technically, he is right, you can calculate it from 2 pieces of info, but you are also right, there is no way to check it.
Kris Morgan, post: 442950, member: 29 wrote: Based on what you've given shouldn't this work?
5729.578/R = D
Then
Delta = L*100*D
Then
((Sin 1/2 Delta)*R)*2
Won't you get your chord length from that?
Did you mean
??=D*L/100
billvhill, post: 443825, member: 8398 wrote: Did you mean
??=D*L/100
LOL! Ya caught the tater. LOL! You're right. Damned algebra.
nm
Kris Morgan, post: 443842, member: 29 wrote: LOL! Ya caught the tater. LOL! You're right. Damned algebra.
Haven't had to use this in a while but it's kind of imprinted in my brain from the past
How do you make the "DELTA" character?
It is one of the symbols on my phone
Kris Morgan, post: 442950, member: 29 wrote: Based on what you've given shouldn't this work?
5729.578/R = D
Then
Delta = L*100*D
Then
((Sin 1/2 Delta)*R)*2
Won't you get your chord length from that?
the chord length still doesn't give you the direction of the chord. tangent-out (azimuth) minus tangent-in should equal the delta as Kris calculated it (or meant to calculate it). If it does, then the the curve is tangent to the two straight lengths (which it should be if they did not provide any additional information.) To get the direction of the chord, you apply ?«Delta to the tangent in.
I usually get the delta by applying the ratio of Delta/360 = L/(2(Pi)R) and solve for Delta. With a calculator you can run that through real easy without having memorized 5729.578. That number used to be used as a short cut when you had to multiply it out long hand.
You can move the 360 to the other side of the = sign, you can also eliminate the 2 and use 180 instead of 360.
?? radians = 180
Just my 2?
Just a shortcut that I use: 18,000 / pi = 5729.5780. That is easier to remember than 5729.5780, at least for me anyway. A long time ago a friend of mine who was a math whiz figured that one out and I have never forgotten it. Carry on.
