We need to figure out an estimated position along an arc using a set of known variables. The starting position on the arc would be called point P, here are the known/measured variables:

P = initial position of point P

x = x coordinate of P (longitude)

y = y coordinate of P (latitude)

V = velocity of point P (nautical miles / min, or nautical miles / hour)

H = heading / directional angle of point P in degrees (000 = North, 180 = South)

R = rate of turn of P in degrees/minute.

All of those variables are known. We actually have a device that is able to calculate point P's rate of turn. See this link here for the device.

We need to calculate the future X,Y positions of point P along the arc based on the variables above with respect to time. The dynamically changing variable to plug into the formula would be Time (in minutes or hours).

For example: at future time T, what would be the position of P (or Px1, Py1) along the arc. My thinking is that it would require some kind of calculus based derivative calculation. Thanks.

  • $\begingroup$ Are you sure $R$ is not the rate of turning of $V$ rather than $P$? If $P$ is the initial position, how does it have a rate of turning? $\endgroup$ – Robert Lewis Sep 20 '17 at 5:16
  • $\begingroup$ P is the antenna on a ship. R is the rate of turn of that ship, which would be the change in the ships gyro heading degrees / minute. So to me it seems it is the rate of turn of both, as V is the velocity of P. As P turns, V is moving with it. $\endgroup$ – wayofthefuture Sep 20 '17 at 15:10
  • $\begingroup$ Not very clear what the data really mean. If $V$ is the velocity as a vector (function of time) then it gives you all necessary information to compute the position starting at $P$. If $V$ is only the speed, but the heading is the instantaneous directional angle, then together they give you all information necessary to compute positions. The need to know $R$ (I presume it is the derivative of $H$) is not needed, unless you only know $H$ initially. $\endgroup$ – H. H. Rugh Sep 21 '17 at 21:15
  • $\begingroup$ We need to compute future positions of P based on the starting point of P. We know the speed, heading, and rate of turn at the initial point P. $\endgroup$ – wayofthefuture Sep 22 '17 at 4:36
  • $\begingroup$ Are $V$, $H$ and $R$ constants? So the point is travelling with constant velocity and simultanously turning by a constant rate $R$? $\endgroup$ – sranthrop Sep 23 '17 at 22:50

Let consider the coordinate $x$ first. Set the origin of your coordinate system at the initial position $P$: the positive direction of the $x$ axis points to East, and time $t = 0$

At any point of time $t$, the $x$-coordinate of your ship will be given by $$ x(t) = \int_0 ^ t v(s) \cos\, \Big(\frac{\pi}{2} - h (s) \Big) \mathrm{d}s$$ where $h$ is the heading (converted to radians), and $v$ is the velocity.

What this hopefully does is get the value of the velocity, compute its projection with respect to the $x$-axis (the $\frac{\pi}{2}$ should take care of the different conventions used for angles, as sailors measure counterclockwise from the $y$ axis), and integrate.

Same for the $y$-coordinate $$ y(t) = \int_0 ^ t v(s) \sin\, \Big(\frac{\pi}{2} - h (s) \Big) \mathrm{d}s$$

You could also consider your point is moving on the complex plane, and wrap both the integrals in $$ z(t) = \int_0 ^ t v(s) \exp \Bigg( i\Big( \frac{\pi}{2} - h (s) \Big) \Bigg) \mathrm{d}s$$ and now your $x$- and $y$-coordinates correspond to $\Re (z)$ and $\Im (z)$, the real and imaginary parts of $z$ respectively.


It seems that my understanding of the proviced information was fallacious. It turns out that velocity is constant, as well as rate of turn. The ship traces an arc of circumference, the radius $r$ of which satisfies the relationship $$ 2 \pi r = v \frac{2\pi}{R}$$ where $R$ is the rate of turn in radians / unit of time. The above equation simply states that the circumference (traced by a ship after a full round) equals the velocity times the time required for a 360 degress rotation.

The ship position vector $X$ as a function of time could be found by the parametric equations

$$X(t) = \left\{ \begin{array}{ll} r \cos (tR + H) \\ r \sin (tR + H) \end{array} \right. $$ to which a easy to figure correction should be added to account for the position $P$ at the beginning of the arc.

  • $\begingroup$ Can you please explain what each of the variables above mean? $\endgroup$ – wayofthefuture Sep 15 '17 at 16:46
  • $\begingroup$ Have added velocity $v$, and switched to $h$ for heading, i think they are all covered now. There are two integrals giving you the $x$ and $y$- coordinate as a function of time, given the functions $v(t)$ and $h(t)$ descriing how velocity and heading vary over time. $\endgroup$ – An aedonist Sep 15 '17 at 16:51
  • $\begingroup$ What is the meaning of the ds? $\endgroup$ – wayofthefuture Sep 15 '17 at 22:46
  • $\begingroup$ @wayofthefuture, $s$ is the Integration variable, and $\mathrm{d}s$ is part of the standard notation for definite integrals, $\int_a^b f(x) \mathrm{d}x$, do you recognise such notation? $\endgroup$ – An aedonist Sep 17 '17 at 21:32
  • $\begingroup$ @wayofthefuture, I appreciate you are not satisfied by my answer: should you wish to point at where it seems wrong / unconvincing to you, I would be glad to improve it. $\endgroup$ – An aedonist Sep 18 '17 at 8:14

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