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From: Michael Daly <michaeldaly_at_home.com> subject: [Paddlewise] Impact loading of tow lines. Date: Sat, 16 Oct 1999 19:22:06 -0400 · This archive was generated by hypermail 2.4.0 : Thu Aug 21 2025 - 16:33:04 PDT

Rather than the usual drivel I spout, I decided to contribute something 
that some of you may find useful.

Some time ago, Julio and others discussed the use of shock absorbing 
materials in towing lines as a means of reducing the impact load on 
the person (or kayak) doing the towing.  Julio tried to calculate the 
forces, but got bogged down by not knowing the formulae to describe 
the physics of the situation.  I wanted to follow up on this, but 
haven’t had the time until now.

The problem is that if a towed kayak is impeded in its motion, the 
sudden rise in tension in the tow line could injure a paddler if 
attached to the tower’s waist.  Alternatively, the force could damage 
the kayak if the line is fixed to it.  In order to analyse this, I have 
made a few assumptions

1. We will look at a worst case scenario - a towed kayak comes to a 
   complete stop while the towing kayak continues.  
2. All energy is absorbed by a bungie in the tow line, none by the tow 
   line itself, the kayak etc.
3. The bungie is a nice, linear-elastic material 

The second assumption makes the resulting forces calculated higher that 
what would occur in reality - this is safe.

The third assumption is iffy.  The bungie material I tested has some 
linear behavior, but I know that real rubber is not - I seem to remember 
that it’s bilinear.  I don't know whether the bungies are made with real
rubber of synthetic.  I’m not sure what the bungie’s real behavior is 
over a wide range of loads. 

The kinetic energy in the towing kayak is what must be absorbed and is 
defined as:

1/2 M V**2  where M = the total mass, V = the velocity and **2 means squared.

The energy absorbed by an elastic material is defined as:

1/2 k x**2   where k = spring constant, x = change in length of the spring

The force in the elastic material is (Hooke’s law)

F = k x    or   x = F/k

Equating the energies:

1/2 M V**2 = 1/2 k x **2  or   M V**2 = k x**2  or  x**2 = V**2 M/k

or  x = V sqrt(M/k).

Since x = F/k,  then   V sqrt(M/k) = F/k 

Therefore    F = V sqrt(kM).

Substituting weight for mass   F = V sqrt( k w/g)  where g = acceleration 
of gravity.

Now we just have to figure out what to do about the k and we can figure 
out the forces and such.

The spring constant, k, is the relationship between the force and the 
change in length of a spring.  You have to determine this experimentally 
for any material.  I rigged up a simple test arrangement in my workshop 
in the basement.  I used some ¼ inch bungie cord and hung it from a ceiling 
via a carabiner and attached weights to the other end to measure the change 
in length over an eight inch section in the middle of the bungie.  By using 
several different weights and repeating this for single, double and triple 
strands of bungie and plugging the numbers into a spreadsheet, I determined 
the elastic constants for each.  The results are a little confusing in that 
two strands of bungie is not exactly twice as stiff as one strand.   Due to 
the degree of stretch and such, I was obliged to use a different weight range 
to calculate the numbers for each of the single, double and triple 
configurations.  I think that the results are revealing some of the 
non-linearity of the material.  

Let us consider a two parallel 1/4 inch bungies.  These have a k value of 
1.6 pounds per inch.  (I regret that, while I am usually more comfortable 
in metric units, for little physics type stuff like this, I am more familiar 
with Imperial units - this reflects the change in units that occurred part 
way through my education.)

Let's assume a paddler weighing 175lb, in a kayak that is 55 lbs carrying 
70 lbs of gear for a total weight of 300 lb.  We’ll also assume a velocity 
of 3 miles per hour.  Plugging the results into our equation (with conversions 
to get everything in inches, pounds and seconds)

3 mph =  3 (5280 * 12)/3600 =  52.8 inches per second.

G = 32.2 ft/sec**2 = 32.2 * 12 = 386.4 in/sec**2

F =  52.8 sqrt[ (1.6 * 300)/ (386.4) ] = 58.8 lb.

This sounds good; 58.8 lb is within the range that some folks considered 
reasonable when it was discussed a while ago.  And that’s a worst case.  
However, this requires the bungie to stretch under this load:

X = F/k = 58.8/1.6 = 36.8 inches!   

Note that this is the _change_ in length.   I’ve noticed that the bungies 
are "comfortable" stretching to twice their length, so this means that the 
bungie has to start with a free length (exclusive of knots) of around a 
yard!   I’ve seen folks using bungies that are only 8 to ten inches long.  
They’d probably reach their limit of stretch before fully absorbing the 
load and then the force would shoot up to a higher level with a bit of 
a shock.

Increasing the stiffness (i.e using three of four strands of bungie or 
heavier bungie) would decrease the stretch, but would increase the force 
as well.  Note that because of the square root in the equation, doubling 
the stiffness only increases the force by 41% (sqrt(2)).  

Anyway, play with your own set of numbers and see what you get.  You can 
assume that three strands of 1/4 inch bungie have a k value of 2.4 and four 
strands are 3.2.   

Mike
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