How to find a few coordinates that are in the straight line, between 2 coordinates?
For example:
Start coordinate: Lat=X1 Long=Y1
End coordinate: Lat=X2 Long=Y2
Make a straight line from X1,Y1 to X2,Y2.
Then find 5 points that are located in that line, that are spread in the same distance.
Anyone can help to find the algorithm and calculation?
The coordinate is in decimal format, e.g. 50.123456, 6.123456
Thanks.
There are no "straight lines" on a sphere (or ellipsoid).
Anyway, you'll need to:
Calculate the distance and initial azimuth from (x1,y1) to (x2,y2).
You can use Vincenty's inverse method.
Calculate the coordinate of points with distance (0,25d, 0.5d, 0.75d) from (x1,y1) at that azimuth (plus points (x1,y1) and (x2,y2) of course).
You can use Vincenty's direct method.
Both direct and inverse methods are described on Wikipedia.
An extremely accurate implementations for both direct and inverse problems are available as part of GeographicLib.
Less accurate, but much simpler methods are described in Aviation Formulary.
Related
I am currently creating a feature and patterning it across a flat plane to get the maximum number of features to fit on the plane. I do this frequently enough to warrant building some sort of marcro for this if possible. The issue that I run into is I still have to manually set the spacing between the parts. I want to be able to create a feature and have it determine "best" fit spacing given an area while avoiding overlaps. I have had very little luck finding any resources describing this. Any information or links to potentially helpful resources on this would be much appreciated!
Thank you.
Before, you start the linear pattern bit:
Select the face2 of that feature2, get the outer most loop2 of edges. You can test for that using loop2.IsOuter.
Now:
if the loop has one edge: that means it's a circle and the spacing must superior to the circle's radius
if the loop has more that one edge, that you need to calculate all the distances between the vertices and assume that the largest distance is the safest spacing.
NOTA: If one of the edges is a spline, then you need a different strategy:
You would need to convert the face into a sketch and finds the coordinates of that spline to calculate the highest distances.
Example: The distance between the edges is lower than the distance between summit of the splines. If the linear pattern has the a vertical direction, then spacing has to be superior to the distance between the summit.
When I say distance, I mean the distance projected on the linear pattern direction.
Say I want to construct a 3D cubic Bézier curve, and I already have both end-points, and the direction (normal vector) for both control points. How can I choose the distance of both control points to their respective end-points in order to make the curve as 'nicely rounded' as possible?
To formalize 'nicely rounded': I think that means maximizing the smallest angle between any two segments in the curve. For example, having end-points (10, 0, 0) and (0, 10, 0) with respective normal vectors (0, 1, 0) and (1, 0, 0) should result in a 90° circular arc. For the specific case of 2D circular arcs, I've found articles like this one. But I haven't been able to find anything for my more general case.
(Note that these images are just to illustrate the 'roundness' concept. My curves are not guaranteed to be plane-aligned. I may replace the images later to better illustrate that point.)
This is a question of aesthetics, and if the real solution is unknown or too complicated, I would be happy with a reasonable approximation. My current approximation is too simplistic: choosing half the distance between the two end-points for both control point distances. Someone more familiar with the math will probably be able to come up with something better.
(PS: This is for open-source software, and I would be happy to give credit on GitHub.)
Edit: Here are some other images to illustrate a 3D case (jsfiddle):
Edit 2: Here's a screenshot of an unstable version of ApiNATOMY to give you an idea of what I'm trying to do. I'm creating 3D tubes to represent blood-vessels, connecting different parts of an anatomical schematic:
(They won't let me put in a jsfiddle link if I don't include code...)
What you are basically asking is to have curvature over the spline as constant as possible.
A curve with constant curvature is just a circular arc, so it makes sense to try to fit such an arc to your input parameters. In 2D, this is easy: construct the line which goes through your starting point and is orthogonal to the desired direction vector. Do the same for the ending point. Now intersect these two lines: the result is the center of the circle which passes through the two points with the desired direction vectors.
In your example, this intersection point would just be (0,0), and the desired circular arc lies on the unit circle.
So this gives you a circular arc, which you can either use directly or use the approximation algorithm which you have already cited.
This breaks down when the two direction vectors are collinear, so you'd have to fudge it a bit if this ever comes up. If they point at each other, you can simply use a straight line.
In 3D, the same construction gives you two planes passing through the end points. Intersect these, and you get a line; on this line, choose the point which minimizes the sum of squared distances to the two points. This gives you the center of a sphere which touches both end points, and now you can simply work in the plane spanned by these three points and proceed as in 2D.
For the special case where your two end points and the two known normal vector for the control points happen to make the Bezier curve a planar one, then basically you are looking for a cubic Bezier curve that can well approximate a circular arc. For this special case, you can set the distance (denoted as L) between the control point and their respective end point as L = (4/3)*tan(A/4) where A is the angle of the circular arc.
For the general 3D case, perhaps you can apply the same formula as:
compute the angle between the two normal vectors.
use L=(4/3)*tan(A/4) to decide the location of your control points.
if your normals are aligned in a plane
What you're basically doing here is creating an elliptical arc, in 3D, where the "it's in 3D" part is completely irrelevant, since it's just a 2D curve, rotated/translated to sit in your 3D space. So let's just solve the 2D case, and then the RT is entirely up to you.
Creating the "perfect" cubic Bezier between two points on an arc comes with limitations. You basically can't create good looking arcs that span more than a quarter circle. So, with that said: your start and end point normals give you a 2D angle between your normal vectors, which is the same angle as between your start and end tangents (since normals are perpendicular to tangents). So, let's:
align our curve so that the tangent at the start is 0
plug the angle between tangents into the formula given in the section on Circle approximation in the Primer on Bezier curves. This is basically just dumb "implementing the formula for c1x/c1y/c2x/c2y as a function that takes an angle as argument, and spits out four values as c1(x,y) and c2(x,y) coordinats".
There is no step 3, we're done.
After step 2, you have your control points in 2D to create the most circular arc between a start and end point. Now you just need to scale/rotate/translate it in 3D so that it lines up with where you needed your start and end point to begin with.
if your normals are not aligned in a plane
Now we have a problem, although one that we can deal with by treating the dimensions as separate things entirely. Instead of creating a single 2D curve, we're going to create three: one that's the X/Y projection, one that's the X/Z projection, and one that's the Y/Z projection. For all three of these, we're going to abstract the control points in exactly the same way as before, and then we simply take the projective control points (three for each control point), and then go "okay, we now have X, Y, and Z projective coordinates. That means we have (X,Y,Z) coordinates", and done again.
I have GPS coordinates provided as degrees latitude, longitude and would like to offset them by a distance and an angle.E.g.: What are the new coordinates if I offset 45.12345, 7.34567 by 22km along bearing 104 degrees ?Thanks
For most applications one of these two formulas are sufficient:
"Lat/lon given radial and distance"
The second one is slower, but makes less problems in special situations (see docu on that page).
Read the introduction on that page, and make sure that lat/lon are converted to radians before and back to degrees after having the result.
Make sure that your system uses atan2(y,x) (which is usually the case) and not atan2(x,y) which is the case in Excell.
The link in the previous answer no longer works, here is the link using the way back machine:
https://web.archive.org/web/20161209044600/http://williams.best.vwh.net/avform.htm
The formula is:
A point {lat,lon} is a distance d out on the tc radial from point 1 if:
lat=asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
IF (cos(lat)=0)
lon=lon1 // endpoint a pole
ELSE
lon=mod(lon1-asin(sin(tc)*sin(d)/cos(lat))+pi,2*pi)-pi
ENDIF
This algorithm is limited to distances such that dlon <pi/2, i.e those that extend around less than one quarter of the circumference of the earth in longitude. A completely general, but more complicated algorithm is necessary if greater distances are allowed:
lat =asin(sin(lat1)*cos(d)+cos(lat1)*sin(d)*cos(tc))
dlon=atan2(sin(tc)*sin(d)*cos(lat1),cos(d)-sin(lat1)*sin(lat))
lon=mod( lon1-dlon +pi,2*pi )-pi
Simmilar with Direction between 2 Latitude/Longitude points in C#
but with objective-c
Also I want a formula that works for large distance near the pole if it's possible.
You'll need the following complete but rather difficult stuff. A slightly easier description is found on wikipedia.
Or you could save yourself a lot of time and use CLLocation's distanceFromLocation method:
distanceFromLocation:
Returns the distance (in meters) from the receiver’s location to the specified location.
Discussion
This method measures the distance between the two locations by tracing a line between them that follows the curvature of the Earth. The resulting arc is a smooth curve and does not take into account specific altitude changes between the two locations.
http://developer.apple.com/library/ios/DOCUMENTATION/CoreLocation/Reference/CLLocation_Class/CLLocation/CLLocation.html#//apple_ref/occ/instm/CLLocation/distanceFromLocation:
I'm trying to figure out how to calculate a min/max lat/long bound on the specific given range of a gps coordinate.
for example: gps coord 37.42935699924869,-122.16962099075317 range .2 miles
I'm looking at the point + range + bearing in the http://www.movable-type.co.uk/scripts/latlong.html site but im not sure if this is exactly what i want.
This gives 4 unique lat/long pairs and I want/need a max/min lat and a max/min long.
Calculate the distance between the (constant) central point and the point you want to test. (This page should give you the distance (in meters)).
If (distance < 0.2) then ...
Well, given a point and a distance, you will get a circle.
You're looking for two points, which will essentially describe a square (two opposite corners). The two points you're looking for won't even be on the circle. I'm not exactly sure why you want this, but I don't think there is an answer to your question.
Perhaps you could tell us what you're trying to accomplish.
EDIT: Added image to illustrate. The orange line is the distance from the centre (e.g. 0.2 miles)
alt text http://img155.imageshack.us/img155/1315/diagramp.png
After your clarification, here is a less elegant answer that might give you what you want. Well, you want the inverse of a really complicated function. I'm afraid my math skills aren't up to the task, but it should be doable.
A less elegant solution is to find it by trial and error. Essentially, keep longitude the same and vary latitude. Using the right algorithm, you should be able to find one that is very close to the distance you want. This will give you a point on the circle (one of four that is also on the square).
Then keep latitude the same and vary longitude. This will give you a second point on the square (on the middle of one of the sides), from there you can find the 4 corners of the square.
This will slow, depending on how often you have to do it, that might or might not matter.