# Thread: Equations for path and position of a Towed Object

1. ## Equations for path and position of a Towed Object

I think this is a problem requiring calculus, but there may be a geometric solution. My skills in these areas are now too rusty.

Suppose you have a vessel (e.g. tug boat) travelling along the x-axis towing an object with a tow rope of length R. Lets set the tug at the origin and the object at (x=-R,y=0). The tug at (0,0) now makes an instantaneous turn to port and heads north east at an angle theta to the x-axis. It travels a distance D in a straight line to arrive at a point T= (D.Cos(theta),D.Sin(theta)). What path did the towed object travel and where is it now (i.e. what x,y)? We know that the towed object is always a distance R from the tug. It does not follow in the wake of the tug but instead follows a curve.

What is the equation that describes the path of the towed object (in terms of the start positions, D and theta)? This problem assumes that the towed object has no mass/momentum and always moves in the current direction of the tow rope.

What happens if the towed object starts not in the wake of the tug and therefore not on the x-axis, e.g. at a point (-R,Y), where Y can be positive or negative ?

Does anyone have an insight into this problem? Have I put this problem in the correct forum section?

Thanks

2. ## Re: Equations for path and position of a Towed Object

Consider the angle between the tow rop and the tug's direction of travel - call this angle theta. The angle theta starts at 90 degrees and as the tug moves forward teh angle assymptoticallly approaches 0 degrees. The differential equation that controls theta is:

$-R \frac {d \theta} {dt} = \frac v R dt$

where 'R' = length of the tow rope and 'v' = velocity of the tug.

This solves as follows:

$\int \frac {d \theta }{\sin \theta} = - \int \frac v R t$

$\ln(\sin(\frac {\theta} 2)) - \ln(\cos(\frac {\theta} 2)) = \frac {-vt}{R} + constant$

Raise both sides by e, and this becomes:

$\tan( \frac {\theta} 2) = C e^{-vt/R}$

$\theta = 2 \tan ^{-1} (e^{-y/R})$

where 'y' is the tug's position. Attached is a plot of this angle versus the tug's position in terms of multiples of R.

3. ## Re: Equations for path and position of a Towed Object

Many thanks for this solution to the towed object problem. I have now implemented them in my software ("GPS Utility") as a mode for offsetting GPS tracks. Although the track of the tug is being recorded, it is the track of the towed object that is more important as this is where data collection is being done. The towed object may be below the water surface and so unable to receive GPS signals. Your equations have enabled a me to provide a better method for correlating sub-sea data with position within my GPSU software. See the attached plot of tug (in blue) and towed object (in red) - a couple of point pairs are shown joined by 'R' to illustrate the tow rope. Thanks again for your help, the solution was beyond my math skills.

4. ## Re: Equations for path and position of a Towed Object

Glad you found it useful - but a word of caution: since you are using this for a real world application please be aware that the mathematical model of an "ideal" tug on smooth water is probably not terribly good at predicting what happens in real life. The mathematical model does not take into account real world conditions such as water currents or the effects of wind on the position of the barge relative to the tug. My guess is that in the real world those effects may overwhelm the effect that Ive modeled for you. This model works pretty well for situations that are more stable - such as modeling the track of the rear wheels of a trailer being towed behind a truck - but I'd be cautious about applying it to a tug and barge on water.

5. ## Re: Equations for path and position of a Towed Object

Thanks for the word of caution. I know that there can be errors, but this model is far better than previous models which assumed that the object travelled in the wake or was on the end of a rigid boom. It will be used mostly for sub-sea objects rather than a surface barge, so the effects of wind and current are reduced. Yes, non-linear currents or currents which differ in depth can be a problem. Other errors can occur if the sub-sea object has steering fins or oscillates from side to side. However for a surface barge, the problem does not arize as it would normally carry its own GPS tracking receiver and the model would not then be required.

Perhaps we should now move this discussion off-line from this maths forum - if you would like to communicate directly then please email me at the address shown on GPS Utility - Service and Support.