Calculating Band Tension for Deadlifts

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Les
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Calculating Band Tension for Deadlifts

#1

Post by Les » Fri Feb 16, 2018 7:49 pm

The title says it all. I'm doing deadlifts on a platform equipped with band pegs. I found this article with a chart in it, but I wasn't sure how it applied to deadlifts. The bands are already stretched a little past 41" just being set up with the bar on the floor. I was using the orange long bands tonight while deadlifting. https://www.elitefts.com/education/moti ... ibrations/

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Re: Calculating Band Tension for Deadlifts

#2

Post by strega » Fri Feb 16, 2018 9:03 pm

That article is way too high brow for me, too much thinking. I bought a fish scale with the simple idea that attach the band and use the hook on the scale and just pull it out and write down the weight. A the very least it's a repeatable process to which you can develop your own standard.

This is the one I own:
Ultimate54 Portable Digital Hanging Hook Fishing and Luggage Scale Multifunction with Tare and Large LED Display & Backlight 110lb/50kg Capacity - Free Luggage Strap Batteries Included Black


One cavaet is I'm not stacking on and using a ton of bands, but so far it's providing me the info I need.

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Re: Calculating Band Tension for Deadlifts

#3

Post by Savs » Sat Feb 17, 2018 6:42 am

Wrong, wrong, wrong.

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Re: Calculating Band Tension for Deadlifts

#4

Post by Savs » Sat Feb 17, 2018 4:29 pm

Everyone, I'm sorry. Just thinking about it now, I realize my post above was wrong. I have to fix it, and that'll take me some time. I copied the earlier post, so I still have that in order to make edits. I won't get to it for a little while and didn't want to leave the mistake. Basically, the tension in the bands doesn't act to pull straight down, so there's an angular correction. Also, while I'm at it, I might as well make it more accurate by calculating the correct equation for each line and not use zero for the y-intercepts as I did before. (I don't even know if anyone cares or followed my earlier analysis, but I'll fix it. Sorry, Les!)

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Re: Calculating Band Tension for Deadlifts

#5

Post by mgil » Sat Feb 17, 2018 4:42 pm

@Savs are you typing this? If you submit it here, just tag me and I’ll move it to the drafts section.

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Re: Calculating Band Tension for Deadlifts

#6

Post by Savs » Sat Feb 17, 2018 4:52 pm

mgil wrote: Sat Feb 17, 2018 4:42 pm @Savs are you typing this? If you submit it here, just tag me and I’ll move it to the drafts section.
Okay, thanks. It might be of general use, but the specific results will be for the EliteFTS bands.

I'll get to scribbling out some stuff and typing it out later this evening. Make some tea, put on some tunes, ... Too bad I don't have one of those #IntellectualLP laptop desks. I'll see what I can do without it. #InvoluntaryHardship

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Re: Calculating Band Tension for Deadlifts

#7

Post by Savs » Sat Feb 17, 2018 10:38 pm

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.

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Re: Calculating Band Tension for Deadlifts

#8

Post by mgil » Sun Feb 18, 2018 4:21 am

@Savs, I’ll still move it since it’s a good start and the discussion can be focused on revision and hopefully getting some data from a few folks if they are able.

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Re: Calculating Band Tension for Deadlifts

#9

Post by Savs » Sun Feb 18, 2018 4:30 am

Come on, you guys. Help! I think the spring constant should be doubled.

If I take a band and stretch it through some distance [math]x[/math], that'll produce some force [math]F[/math]. If I then fold the band in half and stretch the folded-over band through the same distance [math]x[/math], I think the force is [math]4F[/math]. It's quadrupled, not just doubled. The spring constants double and I add the two forces. Or to think of it another way, the individual spring constants double but treating them as one spring, I add them in parallel (which for the spring constants is like adding resistors in series). That produces a multiplicative factor of 4, not 2, in the force.

So, for the example I worked out earlier, at the start of the lift the additional weight is 18 lb, and at the top it's 156 lb. Huh. Les, if you're reading, did it feel that heavy at the top? Anyone wanna do the experiment of folded-over bands?
Last edited by Savs on Sun Feb 18, 2018 4:32 am, edited 2 times in total.

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Re: Calculating Band Tension for Deadlifts

#10

Post by Savs » Sun Feb 18, 2018 4:30 am

mgil wrote: Sun Feb 18, 2018 4:21 am @Savs, I’ll still move it since it’s a good start and the discussion can be focused on revision and hopefully getting some data from a few folks if they are able.
Okay. I just added some more comments in a new post.

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Re: Calculating Band Tension for Deadlifts

#11

Post by Savs » Sun Feb 18, 2018 6:11 am

One more post so I can get some closure on this thing.

For those who didn't the math, I think the data isn't directly very useful. You have to stretch the band by three feet to even get on the chart. It would've been nice if they had data for 0 to 3 feet instead of 3 to 7. So, we have to extrapolate from 3 feet down to a few inches. That's a long way to go, and needs some justifcation. I waved my hands and mumbled something about Hooke's Law.

Also, it would've been nice if they didn't bury in the fine print the fact that the data are for two bands not one.

So, let's assume we know the stretched length at the top of the pull and we have data for te force at that stretched length. We still have to multiply by a factor of [math]2\sin\theta[/math], and that'll take some measuring. We also have to multiply by 2 because the bands are folded over the bar (I think).

I don't know if their data table has ever been used for anything. I really don't know. Maybe somebody used it for the case of one band stretched at least 3 feet pulling directly up or down. Divide the force values by 2, in that case. They calibrated their bands, tho.

For whatever it's worth, with the extra factor of two, the equation for the additional weight [math]W_b[/math] is
[equation]
W_b = 2k(66 + 4h)\left[ 1 - \frac{L_o}{\sqrt{(x_o+L_o)^2 + 4h^2 +132h}} \right]
[/equation]
where the weight is in pounds, the lengths are in inches, [math]k[/math] is the spring constant of the band, [math]x_o[/math] is the amount the band is initially stretched, [math]L_o = 41[/math] is the unstretched length of the band, and [math]h[/math] is the distance the bar is moved.

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Re: Calculating Band Tension for Deadlifts

#12

Post by TimK » Sun Feb 18, 2018 7:39 am

So... looks like the fish scale is the way to go, then?

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Re: Calculating Band Tension for Deadlifts

#13

Post by mgil » Sun Feb 18, 2018 8:04 am

TimK wrote: Sun Feb 18, 2018 7:39 am So... looks like the fish scale is the way to go, then?
To collect data to build the model!

What’s nice about having the model is that you can then calculate the force for any ROM as opposed to guessing or measuring directly.

@Savs when they say “the stretch the band”, is that a single length of elastic or both sides of the loop?

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Re: Calculating Band Tension for Deadlifts

#14

Post by Savs » Sun Feb 18, 2018 8:31 am

mgil wrote: Sun Feb 18, 2018 8:04 am
TimK wrote: Sun Feb 18, 2018 7:39 am So... looks like the fish scale is the way to go, then?
To collect data to build the model!

What’s nice about having the model is that you can then calculate the force for any ROM as opposed to guessing or measuring directly.
Yeah, use a fish scale to get the data for smaller displacements. Put some weight on it, measure the displacement. Get a few data points. I'd be a lot more comfortable with the formula if there were actual data in that range.
@Savs when they say “the stretch the band”, is that a single length of elastic or both sides of the loop?
My guess is both sides. Beats me, though. They say two bands in the fine print, so maybe they mean two single lengths of one band (both sides). Les said, "I found this article with a chart in it, but I wasn't sure how it applied to deadlifts." That makes two of us, Les. :-)

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Re: Calculating Band Tension for Deadlifts

#15

Post by Les » Sun Feb 18, 2018 10:33 am

Thanks for working on this @Savs . :-)

I will try to plug in some different numbers in to the formula and see what I get. I'm guessing it would be both sides of the band. They hooked it up on the bottom of a monolift and checked it from there. For deadlifting, I have both sides on top of the bar, so hopefully that would be the same, except the band is getting stretched a little to be put in position at the start.

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