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I saw some posts on another forum where someone was trying to describe the relationships between the length of the "V" in a towing system and how it related to the actual bridle line tension. So I came up with this little diagram to help explain it.
- Two_Point_Tow_Diagram.png (24.07 KiB) Viewed 5393 times
The diagram shows a tow force being divided by a "V" bridle. Four different bridle lengths are shown as 4 different orange lines passing through 4 different gray rings along the light green "Tow Force" line. The "V" bridle could be oriented in any direction. It could be oriented vertically (maybe between the pilot and the keel) or horizontally (between the pilot's hips or shoulders). It could even be attached to the side of a building since the math is all the same.
It turns out that the relationship between the tow force and the line tension is really based on angles rather than on distances. But I used distances in this formulation because they tend to be easier to measure and visualize. This formulation (based on distances) does introduce an additional arctangent (atan) function, but since the equation requires a cosine anyway, I figured a calculator was going to be part of the process either way.
Whenever you see an equation like this, it's good to test it out in a few simple cases to see if it makes sense. So let's try a couple "extremes".
The easiest extreme to see is when the distance "d" is very long (or "w" is very small). In that case, the two bridle lines are almost parallel, and we would expect that each of the lines would be carrying about 1/2 of the tow force. Let's see if that's what the equation tells us.
When "d" is very long (or "w" is very short), the ratio of w/d gets smaller and smaller ... eventually approaching zero. So does the ratio of w/(2d). So the argument to the arctangent function is approaching zero. If you remember your trigonometry, you'll remember that the arctangent function approaches zero as its argument approaches zero. So that whole "atan()" expression is tending toward zero. Now if you can remember a little more of your trigonometry, you might remember that the cosine function approaches a value of one as its argument approaches zero. So that means the bottom of the fraction is getting to be 2x1 ... which is 2. So the line tension is becoming equal to the tow force divided by two. That's exactly what we would expect.
Now lets look at the other extreme when "d" is very short compared to "w". When this happens, the bridle line tension is expected to go toward infinity. Let's see if that's what the equation predicts.
When "d" becomes very small with respect to "w", the ratio of (w/2d) grows toward infinity. The arctangent approaches 90 degrees as its argument goes toward infinity. So the cosine expression is approaching zero (since the cosine of 90 is zero). Of course, two times zero is still zero, and so we end up with the tow force divided by zero which approaches infinity. In practice, of course, you don't end up with a bridle line tension of infinity because distance "d" won't stay at zero as you apply tension. The line will stretch a little or the attachment points will give a little, and the value of "d" will quickly move from being zero to being some small number. The forces on the line will still be very large, but at least they won't be infinity.
Just to get a feel for actual numbers, I used that formula to calculate the bridle line tension for each of the four "rings" shown in that diagram. I measured the distances in pixels (rather than inches, millimeters, or fathoms) because they're going to become ratios anyway.
I measured the distance "w" between the two tow points as 233 pixels, and I measured the distance "d" for each of the 4 cases as 6 pixels, 30 pixels, 156 pixels, and 447 pixels. When you run those numbers through the equation you'll get this chart:
d1 = 94-88 = 6, so the bridle line tension is about 9.7 times the tow force
d2 = 118-88 = 30, so the bridle line tension is about 2.0 times the tow force
d3 = 244-88 = 156, so the bridle line tension is about 0.63 times the tow force
d4 = 535-88 = 447, so the bridle line tension is about 0.52 times the tow force
So when the "V" is relatively deep (as in the last case with w=233 and d=447), you get a bridle line tension of nearly half (0.52) of the tow force. This is what we expect as each half of the line takes up about half of the force. But when the "V" is very shallow (as in the case with w=233 and d=6), you can get very high bridle line tensions (nearly 10 times the tow force in this example). This has important implications for both the strength of the bridle itself and even more importantly for the breaking strength of any weak links that might be inserted into the "V" portion of the bridle.
By the way, I thought I recalled a trigonometric identity for the cosine of an arctangent, and I found that:
cos(arctan(x)) = 1 / (sqrt(1 + x^2))
So if you like square roots better than cosines and arctangents, this formula is for you.
Notes:
2012/09/27: Thanks to Bill Cummings for suggestions on more consistent use of "bridle line tension" in this description.
