Rigging
Math
(Made
Simple)
A
Primer by
Delbert
L.
Hall, Ph.D.
ETCP
Certified Rigger
ETCP
Recognized Trainer
Lesson
4: Tension on Bridle Legs
Now that we
know how to compute the lengths of the bridle legs, we can compute the
load on
each leg. But
before we do that,
let's discuss the angle between the two legs.
We said in Lesson 3 that bridles are used to create a new hanging
point between two existing hanging points, and this is true. It is also true that the
two bridle
legs share the load being lifted (but not always equally), and bridles
can be
used to reduce the load on hanging points.
You should realize that it is possible to use
different combinations of bridle lengths to get the bridle point at the
same
horizontal position, relative to the to existing hanging points, but at
different
vertical relations to the existing hanging points.
In many cases it is desirable to have the bridle point as
high as possible so that it is not seen.
But, the higher you place this point, the wider the bridle
angle (angle
between the two bridle legs) and the greater the force on the bridle
legs. As a general
rule, the bridle angle
should not exceed 120 degrees. If
the bridle angle is greater than 120 degrees, then the load on at least
one of
the legs will be greater than the load being lifted.
And Òflat bridlesÓ (bridles with very wide angles) can put
tremendous loads on their hanging points.
To compute
the tension on the two bridles, we use the equations:
Tension
on L1 = Load ((L1xH2) / ((V1xH2) +
(V2xH1)))
Tension
on L2 = Load ((L2xH1) / ((V1xH2) +
(V2xH1)))
Before your start screaming,
ÒI canÕt
remember all of that!Ó I
will soon teach you a trick that will make it fairly easy to remember. But before we get to that,
draw out a
diagram like the one below:
Now
in place of the label names, put the following values:
Load
= 500 lbs
L1
= 5Õ
V1
= 4Õ
H1
= 3Õ
L2
= 6.7Õ
V2
= 3Õ
H2
= 6Õ
This will
help you to be able to quickly find the values that you need.
Now,
below is the equation for finding the tension
on Leg 1, but I have color-coded it so that it is divided into three
parts. I have also
drawn colored
arrows in our schematic to correspond to the three parts of our
equation.
Tension
on L1 = Load ((L1xH2) / ((V1xH2)+ (V2xH1)))
Here
is the Òtrick.Ó Since
you want to find the tension on L1, begin by
multiplying L1 (high) by the ÒlowÓ side of Leg 2 (which is H2). If you wanted to
find the tension
on L2, you would multiply L2 (high) by the ÒlowÓ side of the L1
triangle, which
would be H1. Got it? So plugging in these
variables in our
equation we getÉ
Tension
on L1 = Load ((5x6) / ((V1xH2)+ (V2xH1))) or
Tension
on L1 = Load ( 30 / ((V1xH2)+ (V2xH1)))
Next,
we want to
figure out what we divide this number by.
This is actually very easy to remember.
We just need to remember that we multiply V on one side by
H
on the other, and add the two numbers together.
You can also remember that you always
multiply one ÒhighÓ side and one ÒlowÓ side, if that helps you. SoÉ
Tension
on L1 = Load ( 30 / ((4x6)+ (3x3)))
or
Tension
on L1 = Load ( 30 / (24+ 9))
or
Tension
on L1 = Load ( 30 /
33) or
Tension
on L1 = Load ( .909)
Now,
we just plug in the Load and multiply.
Tension
on L1 = 500 ( .909)
Tension
on L1 = 454.5 lbs.
LetÕs
now do Leg 2.
Tension
on L2 = Load ((6.7x3) / ((V1xH2)+ (V2xH1)))
or
Tension
on L2 = Load ( 20.1 / ((V1xH2)+ (V2xH1)))
Did
you notice that the
second and third parts of this equation (the blue and green parts) are
exactly
the same as the equation for finding the tension on Leg 1? Because they are the same,
we do not
have to re-calculate those numbers.
We can plug in the results from our first equation and we
haveÉ
Tension
on L2 = Load ( 20.1 /
33) or
Tension
on L2 = Load (.609)
or
Tension
on L2 = 500 ( .609)
Tension
on L2 = 304.5 lbs.
This
is probably the most difficult problem in this
primer. The most
common mistake
is multiplying when you should divide or dividing when you should
multiply, so
work on keeping those straight.
Work on more problems like this one on the worksheet. After a few problems, you should
get the hang of
how to do them.
