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Rigging math: three-point bridles.

Bridles are very common in many types of theatrical rigging, especially arena rigging, and are used to create hanging points in locations where there are no attachment points directly above from which to hang. The simplest, and most common, type of bridle is a two-point bridle.

A two-point bridle uses two cables (or legs) to create a new hanging point where the two legs join. This junction point is on the same plane as the attachment points for the two legs, just lower and somewhere between the two existing attachment points. The lengths of the two legs of the bridle determine both the height and the vertical position of the bridle point (junction). The Pythagorean theorem is commonly used to calculate the lengths of the two bridle legs.

A three-point bridle is used when the desired bridle point is not directly between two existing attachment points and must be positioned between three points. This is common in spaces where the existing attachment points are scattered about. When the three legs are the proper length, a new hanging point is created that is in the desired location above the stage.


The math for calculating the bridle lengths on a three-point bridle is a little more complicated than calculating the bridle lengths for two point bridles. With two point bridles, you have only three points to be concerned with: the attachment point for each bridle leg and the bridle point where the two legs meet. Since all three of these points are on a single plane, you only need to know the horizontal and vertical distances of the hanging points from the bridle point in order calculate the two bridle lengths. Therefore, you only need to know four numbers. Really simple.

A three-point bridle problem is three-dimensional. You must know the X, Y, and Z coordinates of the attachment points for the three bridle legs and the bridle point. Three coordinates for four points is twelve coordinates numbers. See why it is more complicated?

The first trick to solving this problem is to collect and organize the data. The best way is to complete a coordinates table, such as the one below, with the needed data:
P1 X: __ Y: __ Z: __

P2 X: __ Y: __ Z: __

P3 X: __ Y: __ Z: __

P4 X: __ Y: __ Z: __

Note: P1, P2, and P3 will be the hanging points for our three bridle legs (L1, L2, and L3) and P4 will be the bridle point (where the three legs meet).


Look at the hanging plot (plan view) in figure 1. In this plot you see that the three attachment points are located on two beams, and the bridle point is between then. Each attachment point on the plot, as well as the bridle point, is marked with its X and Y coordinates in parentheses. I called the lower left hanging point on my plot 0,0 so that all of the other coordinate points are positive numbers (since they are in Quadrant I of a Cartesian coordinate plane). You can set up your coordinate system in whatever way makes the most sense for you,

Now that we have the X and Y coordinates shown in figure 1, put them into the table below.
P1 X1: 0  Y1: 0  Z1: __

P2 X2: 0  Y2: 12 Z2: __

P3 X3: 16 Y3: 6  Z3: __

P4 X4: 8  Y4: Z  Z4: __

Next, we need to input the Z coordinates. These are the heights of the points. So, if the bottoms of the two beams are 50 feet above the deck and the bridle point is 35 feet above the deck, the completed coordinate table would look like this:
P1 X1: 0  Y1: 0  Z1: 50

P2 X2: 0  Y2: 12 Z2: 50

P3 X3: 16 Y3: 6  Z3: 50

P4 X4: 8  Y4: Z  Z4: 35

Now that the table is complete, it is time to do some math. The formulas for computing the lengths of the three bridle legs, L1, L2, and L3, are:

L1 = [square root of [(X1 - X4).sup.2] + [(Y1 - Y4).sup.2] + [(Z1 - Z4).sup.2]]

L2 = [square root of [(X2 - X4).sup.2] + [(Y2 - Y4).sup.2] + [(Z2 - Z4).sup.2]]

L3 = [square root of [(X3 - X4).sup.2] + [(Y3 - Y4).sup.2] + [(Z3 - Z4).sup.2]]

This is really just a variation on the Pythagorean theorem. The big difference is that you must subtract the appropriate axis coordinate for the bridle point (P4) from the same axis coordinate for the three attachment-point coordinates before you square it. If you are taking the Entertainment Technician Certification Program (ETCP) exam to become a certified rigger, these formulas are listed on the Formula Table that you will be given (just be sure you are able to recognize them from the many other formulas on the sheet). After you work a few problems, this is actually a pretty easy equation to remember.

So, let's plug in the numbers and calculate the lengths of the bridle legs.

L1 = [square root of [(X1 - X4).sup.2] + [(Y1 - Y4).sup.2] + [(Z1 - Z4).sup.2]]

L1 = [square root of [(0 - 8).sup.2] + [(0 - 7).sup.2] + [(50 - 35).sup.2]]

L1 = [square root of [(-8).sup.2] + [(-7).sup.2] + [(15).sup.2]]

L1 = [square root of 64 + 49 + 225]

L1 = [square root of 338]

L1 = 18.38 feet


L2 = [square root of [(X2 - X4).sup.2] + [(Y2 - Y4).sup.2] + [(Z2 - Z4).sup.2]]

L2 = [square root of [(0 - 8).sup.2] + [(12 - 7).sup.2] + [(50 - 35).sup.2]]

L2 = [square root of [(-8).sup.2] + [(-5).sup.2] + [(15).sup.2]]

L2 = [square root of 64 + 25 + 225]

L2 = [square root of 314]

L2 = 17.72 feet


L3 = [square root of [(X3 - X4).sup.2] + [(Y3 - Y4).sup.2] + [(Z3 - Z4).sup.2]]

L3 = [square root of [(16 - 8).sup.2] + [(6 - 7).sup.2] + [(50 - 35).sup.2]]

L3 = [square root of [(8).sup.2] + [(-1).sup.2] + [(15).sup.2]]

L3 = [square root of 64 + 1 + 225]

L3 = [square root of 290]

L3 = 17.03 feet


There is another way to calculate three-point bridle lengths--one that uses the Pythagorean theorem, the same formula used to calculate the lengths of two-point bridles. The trick to using this simpler equation is that you must use it twice to calculate the length of each bridle leg. Let's look at how this method works.


Figure 2 is an elevation view of Leg 1, the cable between P1 and P4. To compute the length of the bridle leg (L1) we need to know both the vertical and horizontal distances between points P1 and P4. We know that the vertical distance is 15 feet (50 feet minus 35 feet), but we need to figure out the horizontal distance before we can use the Pythagorean theorem to calculate the hypotenuse of this triangle. Fortunately, there is an easy way to find this distance.

The plan view in figure 3 shows the same three hanging points as in figure 1, except I have overlaid the drawing with three right triangles (one green, one red, and one blue). The hypotenuse of each of these triangles correlates with one of the three bridle legs, and the two vertices are on the X and Y axes.

Creating these three right triangles (where the hypotenuses correlates with the bridle legs) is a very important step. Look carefully at figure 3 and understand how these triangles were created. Without these triangles, and knowing the lengths of their vertices, you cannot calculate the horizontal lengths.


Look at the green triangle in figure 3. The hypotenuse of this triangle (the dashed line between P1 and P4) correlates with the horizontal line in figure 2. As you see, the lengths of the two vertices of this triangle are 8 feet and Z feet. By using the Pythagorean theorem, we can calculate the horizontal distance between P1 and P4 (the hypotenuse of this triangle).

H1 = [square root of [8.sup.2] + [7.sup.2]]

H1 = [square root of 64 + 49]

H1 = [square root of 113]

H1 = 10.63 feet

Now that we know the horizontal distance, use this number with the vertical distance (15 feet) to determine the length of the bridle L1.

L1 = [square root of [10.63.sup.2] + [15.sup.2]]

L1 = [square root of 113 + 225]

L1 = [square root of 338]

L1 = 18.38 feet

This is exactly the same length for L1 that we calculated using the first method. As you see, by using this simple formula (twice) we can calculate the length of a three-point bridle.

By using 5 feet and 8 feet for the vertices of the red triangle, the horizontal distance for L2 can be calculated as 9.43 feet. And by using 8 feet and 1 foot for the vertices of the blue triangle, the horizontal distance for L3 can be calculated as 8.06 feet. These numbers, along with the vertical height of 15 feet, can be used to calculate the lengths of L2 and L3.

Both methods for calculating the bridle lengths work equally well. I actually prefer the second method, but you can use the one that fits your way of thinking.


Although calculating the lengths of thee-point bridles is relatively simple, calculating the load on each bridle is more involved. While the math is not complicated, there are several steps. And because each step involves numerous operations that use a lot of numbers, it is very easy to get confused and make mistakes. Since just one mistake can throw off everything, calculating the loads on the bridle legs can be a tedious task, but not an impossible one.

You will, of course, need the bridle lengths that you computed above, but that is just a start. You will need three more formulas (or sets of formulas). Each formula is a "step" in calculating the tension on the legs. In the first step, you will create a formula matrix. This set of formulas calculates nine values that will be needed in steps 2 and 3. Let's get started.


The following matrix is designed to help you remember all nine formulas.
N1X = X1 - X4 / L1   N1Y = Y1 - Y4 / L1   N1Z = Z1 - Z4 / L1

N2X = X2 - X4 / L2   N2Y = Y2 - Y4 / L2   N2Z = Z2 - Z4 / L2

N3X = X3 - X4 / L3   N3Y = Y3 - Y4 / L3   N3Z = Z3 - Z4 / L3

Write out these formulas, substitute the X, Y, and Z values from the coordinate table, and the L dimensions (bridle lengths) which you calculated earlier, and calculate the results. The results of this set of formulas are:
N1X = -.435   N1Y = -.381   N1Z = .816

N2X = -.351   N2Y = .282    N2Z = .846

N3X = .470    N3Y = -.059   N3Z = .881

These results will be used to compute a divisor number in step 2, below. This divisor, D, will be used in the final set of formulas.


The divisor is computed with the following equation:

D = (N1X)(N2Y)(N3Z) + (N2X)(N3Y)(NIZ) + (N3X)(N1Y)(N2Z) - (N3X)(N2Y)(N1Z) - (N2X)(N1Y)(N3Z) - (N1X)(N3Y)(N2Z)

Wow, what a confusing equation to try to remember. Here is a trick that might help you figure out how to recreate this formula without actually remembering all the parts--work on diagonals. Confused? Look at the color-coded version of the results table below.
    N1X = -.435           N1Y = -.381          N1Z = .816

    N2X = -.351           N2Y = .282           N2Z = .846

    N3X = .470            N3Y = -.059          N3Z = .881


D = (N1X)(N2Y)(N3Z)  +  (N2X)(N3Y)(N1Z)  +  (N3X)(N1Y)(N2Z) -
    (N3X)(N2Y)(N1Z)  -  (N2X)(N1Y)(N3Z)  -  (N1X)(N3Y)(N2Z)

Do you see a pattern with the different colors? Look at the three sets of numbers highlighted in blue. They are arranged in a diagonal line from the upper-left corner to the lower right corner. They are your first set of numbers. Replace the variable names in the equation with their values.

Now, look at the group of numbers highlighted in red. They also make a diagonal line, just below the blue line (sure, you have to pick up the stray number at the top of column, but it follows the pattern). They are the second set. Put these values into the equation.

Last, look at the green line of numbers. See the pattern? Put these values into the equation, too.

Now that you have the top line filled in, do the same for the next three groups of numbers--the bottom line. These three diagonal rows move from the bottom left corner to the top right (opposite of the first three groups). See the pattern here?
     N1X = -.435           N1Y = -.381          N1Z = .816
     N2X = -.451            N2Y = .282          N2Z = .846
     N3X = .470            N3Y = -.059          N3Z = .881

D = (N1X)(N2Y)(N3Z)  +  (N2X)(N3Y)(NIZ)  +  (N3X)(N1Y)(N2Z) -
    (N3X)(N2Y)(N1Z)  -  (N2X)(N1Y)(N3Z)  -  (N1X)(N3Y)(N2Z)

By the way, this formula was not included on the last ETCP Formula Table that I saw (but it could be in a newer one). It sure would be useful. But if you remember my visual trick--work on the diagonals--you can create the formula without having to re member what may seem like an interrelated mess of an equation.

Replacing the variables in this formula with the value from the matrix, we get:

D = (-.435)(.282)(.881) + (-.451)(-.059)(.816) + (.470)(-.381)(.846) - (.470)(.282)(.816) - (-.451)(-.381)(.881) - (-.435)(-.059)(.846)

Multiplying these six sets of numbers is probably the most tedious step in computing the load on the bridle legs, and where most people make mistakes. I recommend that you ignore the sign of the sets of numbers as you multiply them (treat them as positive numbers) and apply the sign (positive or negative) after you do the multiplication. Here is how you know the sign of the product (result of multiplying the numbers):

* If there is one negative number in the group, the product is negative.

* If there are two negative numbers in the group, the product is positive.

* If there are three negative numbers in the group, the product is negative.

So, just multiply the three numbers in each of the six groups as though they were all positive numbers, and then apply the appropriate sign. The results of multiplying these groups of numbers are:

D = (-.108) + (.022) + (-.151) - (.108) - (.151) - (.022)


D = -.518

That was a lot of work for this one number, but you will need this number in step 3. You will also need one additional piece of information before you can calculate the load on each bridle leg--the load being supported by the three bridles. For this problem, let's say that the bridles are supporting a 500 pound load. The letter F in the formulas below represents this value. We will call the force (tension) on the bridle legs F1, F2, and F3, respectively.


Here is the final set of formulas:

F1 = ((N2X)(N3Y) - (N3X)(N2Y)) (F / D)

F2 = ((N3X)(N1Y) - (N1X)(N3Y)) (F / D)

F3 = ((N1X)(N2Y) - (N2X)(N1Y)) (F / D)

Fortunately, these formulas are (or were) on the ETCP Formula Table, so you do not have to memorize them. Let's plug in the values and compute the loads on the three bridle legs.

F1 = ((N2X)(N3Y) - (N3X)(N2Y)) (F / D)

F1 = ((-.451)(-.059) - (.470)(.282))(500 / -.518)

F1 = (.026 - .132)(-963.39)

F1 = (-.106) (-963.39)

F1 = 102.12 pounds


F2 = ((N3X)(N1Y) - (N1X)(N3Y)) (F / D)

F2 = ((.470)(-.381)-(-.435)(-.059))(200 / -.518)

F2 = (-.179 - .026)(-963.39)

F2 = (-.205) (-963.39)

F2 = 197.49 pounds


F3 = ((N1X)(N2Y) - (N2X)(N1Y)) (F / D)

F3 = ((-.435)(.282) - (-.451)(-.381))(500 / -.518)

F3 = (-.123-.172)(-963.39)

F3 = (-.295)(-963.39)

F3 = 284.20 pounds


The results of your calculations may vary slightly from the ones above (but not by much--less than one pound) based on how many places to the right of the decimal you extend each number, so do not be confused if your results are not exactly as above.


Calculating the lengths of the three bridle legs was pretty easy, but calculating loads on the three bridle legs was a lot of potentially frustrating work. Computing these loads with a spreadsheet or a rigging app is definitely a lot easier. Still, there are times when you might have to calculate this by hand.

Practice is the only way to become proficient at doing math problems such as these, so create some problems and work them. Be careful; most math mistakes are just foolish errors. Happy calculating.

Delbert L. Hall is president of D2 Flying Effects, an ETCP Certified Rigger, and Recognized Trainer. He is the creator of RigCalc, a rigging app for Android and Apple iOS devices. Delbert is also a professor of theatre and dance at East Tennessee State University and the recipient of United States Institute for Theatre Technology Southeast Regional Section's Founders Award as the Outstanding Educator in Theatrical Design and Technology for 2001. Delbert regularly presents rigging related sessions at the USITT Conference & Stage Expo and at LDI.
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Author:Hall, Delbert
Publication:TD&T (Theatre Design & Technology)
Article Type:Report
Geographic Code:1USA
Date:Jan 1, 2012
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