Hi i am new on strackoverflow, could anyone help me with 2d Terrain trigonometry algorithm, how to achieve something like in screenshot?
this tutorial helps me a lot but still can't figure out how to create mountain terrain
My attempt evolved to this:
Vector (v0.x + t * length, center + amplitude * Mathf.Cos (angle * t));
thanks in advance
it looks like chain so:
generate n random height points
just the peaks covering your screen or map or whatever.
compute chain between neighbors
the chain length should be bigger then the distance between points. For the math see:
Wiki: Catenary
the equation image was taken from the Wiki page linked.
Related
I just cannot figure out how to make an a point with a given velocity move around in cartesian space in my visualization while staying around a sphere (planet).
The input:
Many points with:
A Vector3 position in XYZ (lat/lon coordinates transformed with spherical function below).
A Vector3 velocity (eg. 1.0 m/s eastward, 0.0 m/s elevation change, 2.0 m/s northward).
Note these are not degrees, just meters/second which are similar to my world space units.
Just adding the velocities to the point location will make the points fly of the sphere, which makes sense. Therefore the velocities need to be transformed so stay around the sphere.
Goal: The goal is to create some flowlines around a sphere, for example like this:
Example image of vectors around a globe
So, I have been trying variations on the basic idea of: Taking the normal to center of my sphere, get a perpendicular vector and multiply that again to get a tangent:
// Sphere is always at (0,0,0); but just to indicate for completeness:
float3 normal = objectposition - float3(0,0,0);
// Get perpendicular vector of our velocity.
float3 tangent = cross(normal,velocity);
// Calculate final vector by multiplying this with our original normal
float3 newVector = cross(normal, tangent);
// And then multiplying this with length (magnitude) of the velocity such that the speed is part of the velocity again.
float final_velocity = normalize(newVector) * length(velocity);
However, this only works for an area of the data, it looks like it only works on the half of the western hemisphere (say, USA). To get it (partially) working at the east-southern area (say, South-Africa) I had to switch U and V components.
The XYZ coordinates of the sphere are created using spherical coordinates:
x = radius * Math.Cos(lat) * Math.Cos(lon);
y = radius * Math.Sin(lat);
z = radius * Math.Cos(lat) * Math.Sin(lon);
Of course I have also tried all kinds of variations with multiplying different "Up/Right" vectors like float3(0,1,0) or float3(0,0,1), switching around U/V/W components, etc. to transform the velocity in something that works well. But after about 30 hours of making no progress, I hope that someone can help me with this and point me in the right direction. The problem is basically that only a part of the sphere is correct.
Considering that a part of the data visualizes just fine, I think it should be possible by cross and dot products. As performance is really important here I am trying to stay away from 'expensive' trigonometry operations - if possible.
I have tried switching the velocity components, and in all cases one area/hemisphere works fine, and others don't. For example, switching U and V around (ignoring W for a while) makes both Africa and the US work well. But starting halfway the US, things go wrong again.
To illustrate the issue a bit better, a couple of images. The large purple image has been generated using QGIS3, and shows how it should be:
Unfortunately I have a new SO account and cannot post images yet. Therefore a link, sorry.
Correct: Good result
Incorrect: Bad result
Really hope that someone can shed some light on this issue. Do I need a rotation matrix to rotate the velocity vector? Or multiplying with (a correct) normal/tangent is enough? It looks like that to me, except for these strange oddities and somehow I have the feeling I am overlooking something trivial.
However, math is really not my thing and deciphering formula's are quite a challenge for me. So please bear with me and try to keep the language relative simple (Vector function names are much easier for me than scientific math notation). That I got this far is already quite an achievement for myself.
I tried to be as clear as possible, but if things are unclear, I am happy to elaborate them more.
After quite some frustration I managed to get it done, and just posting the key information that was needed to solve this, after weeks of reading and trying things.
The most important thing is to convert the velocity using rotation matrix no. 6 from ECEF to ENU coordinates. I tried to copy the matrix from the site; but it does not really paste well. So, some code instead:
Matrix3x3:
-sinLon, cosLon, 0,
-cosLon * sinLat, -sinLon * sinLat, cosLat,
cosLon * cosLat, sinLon * cosLat, sinLat
Lon/Lat has to be acquired through a Cartesian to polar coordinate conversion function for the given location where your velocity applies.
Would have preferred a method which required no sin/cos functions but I am not sure if that is possible after all.
I'm studying from a 3D computer graphics book trying to implement my own software renderer for educational purposes. Everything was going great until the book started talking about perspective projection, it got a bit confusing.
The book (Tricks of the 3D Game Programming Gurus) was talking about computing d: the distance from the view point to the view plane. If we look at this scanned image from the book (I hope Andre doesn't mind because I tried contacting him asking but no reply):
Well, here it's clear that the author meant 0.5 * Width besides the figure and not Width.
Then he continues to say:
tan(θ) = opposite/adjacent = y/x, and in this case, y is the distance and x is related to the width.
Looking at the figure, I can't see how this is correct because the opposite of the angle θ/2 is W/2 and the adjacent is d. So:
tan(θ/2) = (W/2) / d =>
d * tan(θ/2) = (W/2) =>
d = (W/2) / tan(θ/2)
Am I wrong? am I missing anything? did the author mean something else?
Any ideas/thoughts/explanations are appreciated.
EDIT: Although in the book's code it does 'work' when multiplication is used, but it's not correct. Because objects should get smaller the higher FOV value is used, that's why division is correct. But if multiplication is used the higher the FOV the bigger objects get. I think the author didn't notice this because he was using 90 degrees FOV angle most the times and tan(90/2)=1 so it doesn't matter if we multiplied or divided.
Okay I'm working on a Space sim and as most space sims I need to work out where the opponents ship will be (the 3d position) when my bullet reaches it. How do I calculate this from the velocity that bullets travel at and the velocity of the opponents ship?
Calculate the relative velocity vector between him and yourself: this could be considered his movement if you were standing still. Calculate his relative distance vector. Now you know that he is already D away and is moving V each time unit. You have V' to calculate, and you know it's length but not it's direction.
Now you are constructing a triangle with these two constraints, his V and your bullet's V'. In two dimensions it'd look like:
Dx+Vx*t = V'x*t
Dy+Vy*t = V'y*t
V'x^2 + V'y^2 = C^2
Which simplifies to:
(Dx/t+Vx)^2 + (Dy/t+Vx)^2 = C^2
And you can use the quadratic formula to solve that. You can apply this technique in three dimensions similarly. There are other ways to solve this, but this is just simple algebra instead of vector calculus.
Collision Detection by Kurt Miller
http://www.gamespp.com/algorithms/collisionDetection.html
Add the negative velocity of the ship to the bullet, so that only the bullet moves. Then calculate the intersection of the ship's shape and the line along which the bullet travels (*pos --> pos + vel * dt*).
The question probably shouldn't be "where the ship will be when the bullet hits it," but IF the bullet hits it. Assuming linear trajectory and constant velocity, calculate the intersection of the two vectors, one representing the projectile path and another representing that of the ship. You can then determine the time that each object (ship and bullet) reach that point by dividing the distance from the original position to the intersection position by the velocity of each. If the times match, you have a collision and the location at which it occurs.
If you need more precise collision detection, you can use something like a simple BSP tree which will give you not only a fast way to determine collisions, but what surface the collision occurred on and, if handled correctly, the exact 3d location of the collision. However, it can be challenging to maintain such a tree in a dynamic environment.
I have to display flight paths on google earth and (still I got the exact flight plan path) want to show in a parabola way (markting side). From the taking off to the landing points.
Basically I'm looking for the math formula to calculate latlng point with altitudes to display the parabola path.
I see how to do the parabolic view with the altitude parameter. But there is less easy, how to get points on a path from two cordinates (start; end).
Thank you !
Maybe I'm wrong about this, but if I remember my physics right, and if the distance between the two points is relatively small compared to the earth's radius, aside from altitude, the path will just follow a great circle.
If you write the coordinates as parametric equations as a function of time, you'd get:
gamma = v_horz/R * t
altitude = g * t * (T0 - t) / 2
(where T0 = 2*v_vert/g = flight time, R = earth radius, g = earth's gravity)
(vertical velocity = derivative of altitude = g/2*T0 - gt = v_vert - gt
where gamma = the angle along the great circle arc followed by the projectile. You know the beginning and endpoints, so you can use spherical trigonometry to figure out the arc distace G0 between start and endpoints. G0 = gamma at time T0 (when projectile lands). This tells you what omega_horz must be (= G0*R / T0).
You can then use spherical trigonometry again to figure out the lat/longitude at any point along the great circle. (use similar triangles -- it's too late in the day for my brain to work through the math properly, sorry)
assumptions:
A.
distance between start/end points is small compared to earth's radius
the flight in question is a ballistic trajectory (some mass M going up and down under the force of gravity alone, no thrust or lift)
we're talking about the planet Earth so you can make certain assumptions for the radius and the force of gravity g
ignore air resistance (good luck in real life)
OR
B.
this is for marketing types so you just want something that looks parabolic, so just use the assumptions in A anyway
EDIT: See also these wikipedia articles on projectile trajectories and great circle distance.
I have a bunch of latitude/longitude pairs that map to known x/y coordinates on a (geographically distorted) map.
Then I have one more latitude/longitude pair. I want to plot it on the map as best is possible. How do I go about doing this?
At first I decided to create a system of linear equations for the three nearest lat/long points and compute a transformation from these, but this doesn't work well at all. Since that's a linear system, I can't use more nearby points either.
You can't assume North is up: all you have is the existing lat/long->x/y mappings.
EDIT: it's not a Mercator projection, or anything like that. It's arbitrarily distorted for readability (think subway map). I want to use only the nearest 5 to 10 mappings so that distortion on other parts of the map doesn't affect the mapping I'm trying to compute.
Further, the entire map is in a very small geographical area so there's no need to worry about the globe--flat-earth assumptions are good enough.
Are there any more specific details on the kind of distortion? If, for example, your latitudes and longitudes are "distorted" onto your 2D map using a Mercator projection, the conversion math is readily available.
If the map is distorted truly arbitrarily, there are lots of things you could try, but the simplest would probably be to compute a weighted average from your existing point mappings. Your weights could be the squared inverse of the x/y distance from your new point to each of your existing points.
Some pseudocode:
estimate-latitude-longitude (x, y)
numerator-latitude := 0
numerator-longitude := 0
denominator := 0
for each point,
deltaX := x - point.x
deltaY := y - point.y
distSq := deltaX * deltaX + deltaY * deltaY
weight := 1 / distSq
numerator-latitude += weight * point.latitude
numerator-longitude += weight * point.longitude
denominator += weight
return (numerator-latitude / denominator, numerator-longitude / denominator)
This code will give a relatively simple approximation. If you can be more precise about the way the projection distorts the geographical coordinates, you can probably do much better.
Alright. From a theoretical point of view, given that the distortion is "arbitrary", and any solution requires you to model this arbitrary distortion, you obviously can't get an "answer". However, any solution is going to involve imposing (usually implicitly) some model of the distortion that may or may not reflect the reality of the situation.
Since you seem to be most interested in models that presume some sort of local continuity of the distortion mapping, the most obvious choice is the one you've already tried: linear interpolaton between the nearest points. Going beyond that is going to require more sophisticated mathematical and numerical analysis knowledge.
You are incorrect, however, in presuming you cannot expand this to more points. You can by using a least-squared error approach. Find the linear answer that minimizes the error of the other points. This is probably the most straight-forward extension. In other words, take the 5 nearest points and try to come up with a linear approximation that minimizes the error of those points. And use that. I would try this next.
If that doesn't work, then the assumption of linearity over the area of N points is broken. At that point you'll need to upgrade to either a quadratic or cubic model. The math is going to get hectic at that point.
the problem is that the sphere can be distorted a number of ways, and having all those points known on the equator, lets say, wont help you map points further away.
You need better 'close' points, then you can assume these three points are on a plane with the fourth and do the interpolation --knowing that the distance of longitudes is a function, not a constant.
Ummm. Maybe I am missing something about the question here, but if you have long/lat info, you also have the direction of north?
It seems you need to map geodesic coordinates to a projected coordinates system. For example osgb to wgs84.
The maths involved is non-trivial, but the code comes out a only a few lines. If I had more time I'd post more but I need a shower so I will be boring and link to the wikipedia entry which is pretty good.
Note: Post shower edited.