# Hyperbola | Eclipse RCP Example | Code Editor library

## kandi X-RAY | Hyperbola Summary

## kandi X-RAY | Hyperbola Summary

Eclipse RCP Example

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### Top functions reviewed by kandi - BETA

- Create the TranscriptControl
- Sends a message
- Process a message
- Render the message
- Create the tree control
- Gets the users
- Initialize drag and drag operation
- Create the dialog area
- Initialize the users
- Stop the workbench
- Create the buttons for the dialog
- Fill the toolbar
- Start the application workbench
- Convert a presence to its key
- Override this method to display the user pressed OK
- Load connection descriptors
- Returns an image for the specified element
- Handle drop target event
- Fill hyperbola bar
- Compares two ChatEditorInput
- Sets the details of the Hyperbola login
- Create the actions for the given workbench window
- Ok for the OK button
- Create the initial layout
- Demonstrates how to add a contact
- Initialize the window

## Hyperbola Key Features

## Hyperbola Examples and Code Snippets

## Community Discussions

Trending Discussions on Hyperbola

QUESTION

I am writing a 2D simulator and game using the HTML canvas which involves orbital mechanics. One feature of the program is to take the position and velocity vector of a satellite at one point and return the semi-major axis, eccentricity, argument of periapsis, etc of a 2D orbit around one planet. When the eccentricity is less than one, I can easily graph the orbit as an ellipse using ctx.ellipse(). However, for eccentricities greater than one, the correct shape of the orbit is a hyperbola. At the moment, my program just draws nothing if the eccentricity is greater than one, but I would like it to graph the correct hyperbolic orbit. Since there is no built in "hyperbola" function, I need to convert my orbit into a Bézier curve. I am at a bit of a loss as to how to do this. The inputs would be the location of one focus, semi-major axis, eccentricity, and argument of periapsis (basically how far the orbit is rotated) and it should return the correct control points to graph a Bézier curve approximation of a hyperbola. It does not have to be exactly perfect, as long as it is a close enough fit. How can I approach this problem?

...ANSWER

Answered 2021-May-12 at 15:45In terms of conic sections, hyperbola are unfortunately the one class of curves that the Canvas *cannot* natively render, so you're stuck with approximating the curve you need. There are some options here:

- Flatten your curve, by sampling the hyperbola at one or two points in the distance and
*lots*of points near the extrema so that you can draw a simple polygon that*looks*like a curve. - Model the hyperbola with a single "best approximation" quadratic or cubic curve.
- As @fang mentions: sample the curve at a few points and convert the Catmull-Rom spline through those points to Bezier form.
- Combine approaches 1 and 2. using a single Bezier curve to approximate the part of the hyperbola that actually looks curved, and using straight lines for the parts that don't.
- Combine approaches 1 and 3, using a Catmull-Rom spline for the curvy bit, and straight lines for the straight bits.

Curve flattening is basically trivial. Rotate your curve until it's axis-aligned, and then just compute `y`

given `x`

using the standard hyperbolic function, where `a`

is half the distance between the extrema, and `b`

is the semi-minor axis:

QUESTION

I have an input between 0 and 1, inclusive, which produces a linear result. I need to convert this to a hyperbola curve, where the result is also between 0 and 1, inclusive. This should produce a rapidly ascending value at first, then slowly increasing to 1 at the end.

I'm coding this in Swift, but really I am looking for the formula (or pointers) and I don't know where to start.

...ANSWER

Answered 2020-Dec-15 at 14:15You can apply any exponent to a value between 0 and 1 without changing the range. So you probably could do something of that kind easily, for instance by using a power to a curve whose the two axis are inverted (or any other formula maybe closer to your exact needs) :

QUESTION

I have the datafile:

...ANSWER

Answered 2020-Nov-11 at 10:51I don't know the algorithm used by gnuplot. Probably an iterative method starting from guessed values of the parameters. The difficulty might come from not convenient initial values and/or from no convergence of the process.

From my own calculus the result is close to the values below. The method, which is not iterative and doesn't require initial guess, is explained in the paper : https://fr.scribd.com/doc/14674814/Regressions-et-equations-integrales

FOR INFORMATION :

The linearisation of the regression is obtained thanks to the integral equation

to which the function to be fitted is solution.

The paper referenced above is mainly written in French. It is pratially translated in : https://scikit-guess.readthedocs.io/en/latest/appendices/references.html

QUESTION

I have 3 receivers (A, B and C), and some signal producing source (let's say sound or light) with an unknown location. Given the locations of A,B and C, and the time at which each receiver "heard" the signal, I'd like to determine the direction of the source.

I understand there are ways to do so with TDoA multilateration/trilateration, however I'm having trouble implementing the calculation. There isn't a lot of clear, detailed information on this out there for those entirely new to the subject. What is out there is vague, more theoretical, or a bit too esoteric for me.

Some similar posts on SO (but not quite what I'm after): TDOA multilateration to locate a sound source Trilateration of a signal using Time Difference(TDOA)

This is also interesting, but assumes we have some boundaries: Multiliteration implementation with inaccurate distance data

@Dave also commented an excellent and fairly accessible resource https://sites.tufts.edu/eeseniordesignhandbook/files/2017/05/FireBrick_OKeefe_F1.pdf, but it falls short of going into enough depth that one might be able to actually implement this in code (at least, for someone without deep knowledge of regression, finding the intersection of the resulting hyperbolas, etc).

[EDIT]: I should add that I can assume the 3 sensors *and* the source are on the surface of the Earth, and the effects of the curvature of the Earth are negligible (i.e. we can work in 2-dimensions).

ANSWER

Answered 2020-Sep-09 at 15:25The simplest (but not fastest) approach would be to solve the equations with gradient descent.

I'm assuming that we know

- the positions of the receivers, A, B, and C, which do not lie on the same line;
- the pseudorange of the unknown source X to each of A, B, and C.

Intuitively, we simulate a physical system with three ideal springs configured like so, where the equilibrium length of each spring is the corresponding pseudorange.

QUESTION

ANSWER

Answered 2020-Jul-06 at 21:38Use of `lines`

creates a continuous lines by joining the points. The two functions are for the upper and lower portions so they both connect points (-1, 0) and (1, 0).

There are likely other ways to accomplish this, but the changes below show what's happening:

QUESTION

I read Feynman's Lecture on Physics Chapter 9 and tried to my own simulation. I used Riemann integrals to calculate velocity and position. Although all start-entry is same, my orbit look's like a hyperbola. Here is lecture note: https://www.feynmanlectures.caltech.edu/I_09.html (Table 9.2)

...ANSWER

Answered 2020-Jun-14 at 05:51`dt`

is supposed to be infinitly small for the integration to work.

The bigger `dt`

the bigger the "local linearization error" (there is probably a nicer mathematical term for that...).

`0.1`

for `dt`

may look small enough, but for your result to converge correctly (towards reality) you need to check smaller time steps, too. If smaller time steps converge towards the same solution your equation is `linar enough`

to use a bigger time step (and save comptation time.)

Try your code with

QUESTION

I need to plot an hyperbola of the type (y-a*x)(y-b*x)=-1 with matplotlib, I've tried to express y(x) but it gives an sqrt with impossible values. How can I assign to y the values to be plotted?
The function i want to plot is

$$(y-\log(r^a 10^c))(y-\log(r^b10^d))=-1$$

which I "can" be rewritten as

$$y=1/2[log(r^(a+b)10^(c+d))\pm\sqrt{-4+\log^2(r^(a-b)10^(c-d))}]$$

ANSWER

Answered 2020-Feb-19 at 14:34Community Discussions, Code Snippets contain sources that include Stack Exchange Network

## Vulnerabilities

No vulnerabilities reported

## Install Hyperbola

You can use Hyperbola like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the Hyperbola component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .

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