Les, the calculation is a goddamn mess. I worked on it for a few hours this morning, screwed that up, and worked on it for a few more hours just now. What I'm saying is I think I now know why EliteFTS didn't write something for bands on deadlifts.
To start, we'll make some approximations so the calculation may be off by a few percent because of those. For example, I don't know the distance from the platform to the pegs. The band wraps around the pegs and goes over the bar, so that'll throw off the measurements a little, too. I don't know how much you initially stretched the bands. I'll leave it as a variable, but will use 4" to get some numbers at the end. Another approximation we'll use is to say the band is acting like a spring, that is, it follows Hooke's Law. A rubber band doesn't follow Hooke's Law as well as a spring does, but it's close enough over a small range of motion.
Hooke's Law is linear. That is, if we plot the force [math]F[/math] of the band versus the stretched length of the band, we should get a straight line. Another way to say it is the force is directly proportional to the amount we stretch the band. We write that relationship using a constant of proportionality we call the spring constant, [math]k[/math], so [math]F = - k \, x[/math] where [math]x[/math] is the amount the band is stretched and the minus sign indicates that the force acts opposite to the direction of stretching. Moving forward from here, we'll drop the minus sign because we can.
There are two orange bands in your link, and I don't know which one you used. Let's get equations for all of them just for future reference. Maybe others can use the results directly if they use EliteFTS bands. For different bands, one would have to get some data by stretching the bands and measuring the force, perhaps using a method strega suggested.
Using the data in your link, we first calculate the spring constant for all of the bands. Their data don't perfectly agree with Hooke's Law, but the results are close enough. The data show the force (in pounds) for an
amount of stretch (in inches) for
two bands (that it's for two bands is in the small print at the bottom). So all of this analysis will be for two bands. If one #band is used, divide the equation at the end by 2.
To calculate the spring constant, we need the slope of the line. That is, we need the first derivative. Ignoring the minus sign, we get [math]k = \Delta F / \Delta x[/math], where [math]x[/math] is the amount of stretch. [math]\Delta[/math] means a change in the value of the variable it precedes, so [math]\Delta x[/math] for example means the change in stretched length. Using data for stretched lengths of 40" and 60" ([math]\Delta x = 20[/math]), I get the following spring constants for the bands.
Micro Mini: [math]k = 0.35[/math] lb/in
Mini: [math]k = 0.62[/math] lb/in
Monster Mini: [math]k = 1.1[/math] lb/in
Light: [math]k = 1.5[/math] lb/in
Average: [math]k = 2.3[/math] lb/in
Strong: [math]k = 3.6[/math] lb/in
Now, there's a little problem with trying to get a linear fit to the data. If we use the point slope equation for a line, we can write down linear equations for the data for each band using the slopes we just calculated. We're supposed to get [math]F = k \, x + b[/math], where [math]b = 0[/math] for Hooke's Law. We don't get [math]b = 0[/math] for any of the equations, although a few of them are close. The equations, dropping the units (the force [math]F[/math] and therefore the values of the [math]b[/math]s are in pounds) and using the slopes between the two points mentioned above, are
Micro Mini: [math]F = 0.35x + 1.5[/math]
Mini: [math]F = 0.62x + 3.7[/math]
Monster Mini: [math]F = 1.1x + 0.8[/math]
Light: [math]F = 1.5x + 4.2[/math]
Average: [math]F = 2.3x + 6.5[/math]
Strong: [math]F = 3.6x - 2 \, .[/math]
If we keep [math]b[/math] in our equations, it'll turn out that we'll compare [math]b[/math] with [math]kL_o[/math], remembering [math]L_o = 41[/math], and in most cases [math]b \ll kL_o[/math]. We'll therefore set [math]b = 0[/math] in all the equations, because soon things will be messy enough.
Moving along, we next calculate how much downward force is on the bar. We'll call the downward force the weight [math]W_b[/math], where the subscript is to remind us it's the extra weight due to the influence of the bands. The bands are in tension and exert a force [math]F[/math] on the bar at angle [math]\theta[/math] ("theta") away from horizontal. As shown in
this diagram, [math]W_b = 2 F \sin\theta[/math].
Next we
draw a diagram to show the side view of a band and the bar on the platform. As shown in the diagram, [math]\sin\theta = (33 + 2h)/(x + L_o)[/math], where the units are in inches. [math]x[/math] is the amount the band is stretched, [math]L_o = 41[/math] is the unstretched length, and [math]h[/math] is the distance the bar is moved. The beginning of the lift corresponds to [math]h = 0[/math] and the end of the lift might be something like [math]h = 12[/math] or 15 or whatever.
Combining all our equations so far, we get
[equation]
W_b = kx \left(\frac{66 + 4h}{x + L_o} \right) \, .
[/equation]
The problem is [math]x[/math] is a function of the distance [math]h[/math] the bar is raised and the amount of initial stretch which we'll call [math]x_o[/math]. As shown in the previous figure, we can write [math]x = \sqrt{(x_o + L_o)^2 + 4h^2 + 132h} - L_o[/math] so our equation for the added weight becomes
[equation]
W_b = k(66 + 4h)\left[ 1 - \frac{L_o}{\sqrt{(x_o+L_o)^2 + 4h^2 +132h}} \right] \, .
[/equation]
We can simplify it a little when the initial stretch is small ([math]x_o \ll L_o[/math]) and for a range of motion [math]h \leq 15[/math], but the result isn't much prettier. So, that's it, unless it's wrong! Choose the [math]k[/math] for the band, plug in [math]L_o = 41[/math], enter a value for [math]x_o[/math], the initial stretch, choose a value for [math]h[/math], the distance the bar is moved, and bust out the calculator. Or plot the added weight as a function of [math]h[/math]. I don't feel like doing either one. What is simpler to do is to calculate the weight at the start ([math]h=0[/math]) and at the end (let's say [math]h =15[/math] for Les). If Les stretched the band by 4" to get it over the bar and on the pegs, and used the "Light" orange band, at the start of the lift the additional weight was 9 lb. At the end of the lift (using 15" for the range of motion), the additional weight was 78 lb.
I suppose this doesn't rise to draft status because it's such a mess. If anyone finds errors, I'm sure you'll point them out. I'm wondering about the tension and the force shown in the first diagram. I think it's okay, though. Les, maybe you're sorry you asked.
Oh, I should add one more thing. The distance from the bar to the pegs, and therefore the initial angle the bands make with the horizontal, doesn't matter for the calculation. All you need to know is the band's "spring constant," unstretched length, and the initial stretch.
ETA: In the last paragraph, I said the angle drops out. That is, it doesn't matter. That's poorly worded. The angle is still in there, but doesn't appear explicity. I deleted that sentence. Also, I'm still wondering about the force as mentioned in the paragraph just above the last one. I was thinking that the band doesn't slip over the bar, so one band becomes two. The bands are cut in half so the spring constants double for each band. Are there any engineers here who've made it this far and want to comment on whether the spring constant in the formula should be doubled? That would double the added weight. I'm not combining springs, so no worries about whether they're added in series or parallel.