Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 8 years ago.
This question appears to be off-topic because it lacks sufficient information to diagnose the problem. Describe your problem in more detail or include a minimal example in the question itself.
This question does not appear to be about programming within the scope defined in the help center.
Improve this question
I have a problem it is like trig and pre calc. We are given vector points for OA which is (6 0) and vector OB (-6 0) and OC (5 root11) and they are all points on a circle with a radius of 6. How exactly do you determine or show that these points lie on the circle?
For any three points which are not mutually co-linear, there is a point equidistant from each member of the set. Finding this point given three arbitrary points (and not given the radius) is a much more informative project if you're learning trigonometry.
In your example, the solution is simple: x2 + y2 = r2. Each of the points you provided have values such that the sum of their squares equals 36, as expected for a circle of radius 6. Your problem is especially easy because the circle in question is already centered on the origin.
Related
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 7 years ago.
Improve this question
I have a camera I'd like to mount on a tripod. I know the FOV dimensions of the camera (22.5 deg V 31 deg H). I'd like to know at what height and what tilt angle to place the camera to be able to capture the two points of interest on the ground. Please see the figure below.
I have attempted a solution using basic trig but am not sure it is correct. Please help!
Your calculations are correct in my view.
If you have to include both the red points, you need to place your camera at an angle such that it encloses both the red points.
So, tan θ = d/hc
=> θ = tan -1 d/hc.
But, as you need to enclose both the points, your angle should be slightly greater than θ ( my suggestion --- some θ+ε, where 0.25 < ε < 1 ) to perfectly capture those 2 red dots.
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
Closed 8 years ago.
This question appears to be off-topic because it lacks sufficient information to diagnose the problem. Describe your problem in more detail or include a minimal example in the question itself.
Questions concerning problems with code you've written must describe the specific problem — and include valid code to reproduce it — in the question itself. See SSCCE.org for guidance.
Improve this question
I have no much time to research net, so can somebody derive the formula for calculating the point belonging to the circle?
as u can see we have point 5,-5 and center, and radius and angle. Thanks for advance, i know that isnt programming problem, but I need this to my project. OFC I see thats point 1,-1. but my pc probably will not ;/
For points inside the circle with radius R and center X,Y:
point (x,y) is inside the circle if:
(X-x)^2+(Y-y)^2 < R^2
Is that what you're after? No.
So what you need is a translate - rotate - translate.
So X,Y is the centre, x,y is the point and angle is the angle in degrees
# Translate
x = x-X ; y = y-Y
# Rotate
rad = angle*M_PI/180
xr = x*cos(rad) + y*sin(rad)
yr = -x*sin(rad) + y*cos(rad)
x = xr+X ; y = yr +Y
So if you only want 90 degrees, replace cos(rad) with 0 and sin(rad) with 1.
Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 9 years ago.
Improve this question
I have searched a lot of google, reviewed my old mathematics books, But I am stuck why can't we find area of an irregular shape from its perimeter? If we have find perimeter from area, Is that possible to reverse the process and find area from perimeter.
This process is working fine for square and rectangle but not for totally irregular shapes.
I'm asking this question because we have one field with irregular shape, and we want to know its area, We can measure its 5 sides.
Don't we have any method to find area from perimeter for irregular shapes.
Any help will be appreciated.
Consider a rope and form an irregular shape. Measure it's area. Now change the shape you just formed with the rope. Measure the area again. Is it the same? Not necessarily.
Or consider a square and a rhomb (which is not a square) with the same perimeter. Do they have the same area? No. Therefore, how do you think could you find out what its area is if you don't have the information of the shape?
You cannot get the area if you just have the information of the perimeter, but don't know anything else.
Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 9 years ago.
Improve this question
I would realize the papercraft of a geodesic sphere like THIS .
Is the dual of the one discussed in THIS OTHER QUESTION .
Where can I find the printable nets, or the online resources to create them?
In the other discussion I learned that there are 3 classes of possible tessellations.
One of these classes needs less pentagons to tessellate the sphere?
I don't exactly know how much faces I need, indicatively between 100 and 200.
Which chord factors I should consider?
Thanks to all
genna,
The link that I provided in the former discussion...
http://thomson.phy.syr.edu/thomsonapplet.php
...provides for display of a dual. Unfortunately, the java applet will only export a data set for the geodesic triangular tessellation; not the dual.
The dual can be constructed from the triangular tessellation, by importing the vertex (point) data set (from the java applet) into a 3D-graphics program, and then following the rules of the definition of a dual. (Each pent/hex face will use the triangular vertex point as the centerpoint of a circle, being equidistant from the sphere center and normal to the central vector. The intersections of these circles will produce the linear edges of the dual faces.)
All tessellated duals, based on the icosahedron, will have the same number of pentagonal faces (12), regardless of how many hex faces are produced.
-Taff
Closed. This question is off-topic. It is not currently accepting answers.
Want to improve this question? Update the question so it's on-topic for Stack Overflow.
Closed 11 years ago.
Improve this question
I want to draw a 3d line from one 3d point to another 3d point.
I know the source point, but need to calculate the destination point. I have the angle and the length of the line. The y coordinate (the height) is the same for the source and the destination point.
Thanks
Basic trig, imagine a bunch of triangles along the various planes and solve for the sides/angles you want to know.
If you give us some sample values for what you already know we can show you the maths.