Welcome to part two to my series on baffles, optimization. Where my thread last week was meant to dispel some common misconceptions on shape, this thread will be (mostly) shape-agnostic, and instead focus on how to optimize the different parameters (baffle height, spacing, etc.) on your choices of fabric and down to get the lowest possible weight (or cost) for a given level of insulation. That said, for my examples I will be using the truncated circle shape from the previous thread, since it is relatively mathematically simple and easy to manipulate, and makes a useful reference point. However, I do provide the tools necessary for elliptical baffles at the end of the post.

Overall, optimization is a simple process to understand. We need to pick a parameter and find even the minimum or the maximum for a given range of inputs. In our case, we are looking for themaximum warmth to weight ratio. To do this, we'll need equations for shell weight, down weight, and baffle wall weight, along with an equation for calculating the R-value. I'll go through the process for deriving them all below.

The one limit I recommend on shapes is that they you limit yourselves to a "truncated" version of some shape. That is, instead of using a baffle shape of a rectangle with half an ellipse or oval on the top and bottom, you use a longer ellipse or oval with the ends cut off. This is a more accurate way to think of baffle shapes, and should provide you with better, more interesting results -- though it might take you a bit to get a feel for the shape. Of course, the capped rectangle shape is easy enough to optimize as well, should you so desire. My guide also assumes the shapes are centered at 0,0 on the Cartesian plane.

But first, aquick note on if you should even bother: In general,the gains from optimizing are very small. The "sweet spot" for baffle dimensions at a given thickness is pretty generous, and the rules of thumb most people follow get you within 5% in most cases, often closer.You'll be shaving maybe 0.25-0.50 oz off a 40F quilt, perhaps an ounce off a 20F if you are lucky. Whether or not that is worth the effort is more of a personal question, but it probably isn't something most people should bother with. But if you are the kind of person who tries to make the lightest possible quilt, this is a good place to start.

Although the utility of the optimization process is pretty low, going through the process was pretty useful, and taught me a fair amount. After all, its not as if anyone knew beforehand exactly how much there was to be gained or lost before I started deriving this stuff (at least, not that I could find). Just determining that made this undertaking useful.

1. Parameters

Before we craft our equations, it is necessary to settle on what parameters we will be using for our equations. This will vary based on your choice of shape and personal preference. For our truncated circle shape, we will need only the radius (r) to define the shape itself, along with a length (x) to represent both how much of our circle is cut off, as well as the baffle spacing. I will be using x to represent half the actual baffle length, just because it makes things a little easier. More complex shapes will need more variables (for example, an ellipse would need two radii, and any differential baffle would require a theta or similar parameter). We will also need the weights for different fabrics used to construct the baffle, these will be detonated withWsfor shell weight andWwfor baffle wall (we can use a simple average of the inner and outer shell weights, as they are equal lengths in our test case). Finally, we need the fill power of the down (F). These variables should be sufficient for all our equations. I recommend oz/yd for the weights, as it makes it easier to interface with the fill power. Likewise, the radius and width will be in inches.

2. Equations

To sort this out, we are going to need a pretty intimidating-looking equation. But when you break it down into smaller pieces, it is actually pretty straightforward. The pieces are as follows:

2a. Equation for Box Baffle Shell Weight

For this, we need to find the perimeter of our shape (excluding the baffle walls) and multiply it by the fabric weight. For the truncated circle, this is easy. The perimeter is calculated using the equation for arc length. I'll jump straight to the good stuff.

Box Baffle Shell Weight = (4*ArcSin[x/r]*r)*(Ws/1296)

Note that we divide the shell weight by 1296 to put it in oz/in

2b. Equation for Box Baffle Wall Weight

For this, you only need the equation for your shape. Plug in length plugged in for x, and you'll have the amount of material, and from there you only need to multiply by the fabric weight. For our circle, it is:

Box Baffle Wall Weight = 2*Sqrt[(r^2-x^2]*(Ww/1296)

We multiply by two in order to capture the half of the baffle wall below the x axis. We only need to account for one wall per baffle, because a baffle shares its walls with its neighbors.

2c. Equation for Fill Weight

For this, we need to determine the area of our shape, and divide by the fill power. This is where calculus starts to come into play. You'll need to determine the definite integral of your shape from -x to x — again, not forgetting any parts that fall beneath the x-axis. For a circle, we wind up with the final equation:

Fill Weight = 2*(Sqrt[r^2 - x^2]*x + r^2*ArcSin[x/r])/F

2d. Equation for R value

I explained how to derive this in my last post, so I won't rehash it here. It involves calculus and harmonic means.

R-Value =(2*x*3.93)/(ArcSin[x/r])

As a side note, I suspect that the R-value I have for down is on the low side. Anyone have more accurate R, U, or k values for down?

2e. Final Equation

In order to determine the the R-Value per unit of weight, we need to combine all our previous equations together. Too keep things clean, I'll use the names of the equations instead of their actual values. The process is to divide the R-value for the baffle by the sum of the different weights. In order to make comparisons between baffles easier, we will also divide the baffle weights by the length, to get the weight per length. When you simplify, you wind up with this equation:

R-Value/Weight ratio = (2*x*R-Value)/(Box Baffle Shell Weight + Box Baffle Wall Weight + Fill Weight)

3. Optimizing

Alright, so now we can calculate the warmth-to-weight ratio of a given baffle. How do we go about determining the minimum possible value over a range of inputs? There are two ways to determine this. The first way is to do some calculus to determine local minimums and maximums. Now, I love math, but that is a bit much even for me. The second way is to make a computer do it for you, and, dear reader, this is what I did. I used Mathematica, since I know it well. You can use the cloud based version for free if you follow this link. Simply copy and paste the code below, then hit ctrl-enter:

And then, on a new line, input the values for the different variables to the FindOptimumBaffle function. For example, say you had a 0.67 oz inner shell and 1.1 oz outer shell (so 0.885 oz average), 0.67 oz noseeum mesh for baffle walls, 750 fill down, and were aiming for a 3" center-of-baffle height. Then you would type FindOptimumBaffle[0.885,0.67,750,1.5], and again hit ctrl-enter. This will return the highest ratio, and the x value that provided it. In our example, it returned {2859.26,{x->1.02918}}. This translates into a quilt with ~2.06 inch wide and 3 inch tall baffles, with a baffle wall height of ~2.18.Code:BoxBaffleShellWt[ShellWt_,r_,x_] := (4*ArcSin[x/r]*r)*(ShellWt/1296) BoxBaffleWallWt[BaffleWt_,r_,x_]:=(2*Sqrt[r^2-x^2])*(BaffleWt/1296) BoxBaffleFillWt[FillPower_,r_,x_]:=(2*(Sqrt[r^2 - x^2]*x + r^2*ArcSin[x/r]))/FillPower BoxBaffleR[r_,x_]:=(2*x*3.93)/(ArcSin[x/r]) BoxBaffleRatio[ShellWt_,BaffleWt_,FillPower_,r_,x_] := BoxBaffleR[r,x]/((BoxBaffleShellWt[ShellWt,r,x]+BoxBaffleWallWt[BaffleWt,r,x]+BoxBaffleFillWt[FillPower,r,x])/(2*x))

You can then plug the values into your R-value function to get the R-value of the baffle. Again, in our example, it would be BoxBaffleR[1.5,1.02918]. This is a little less elegant than inputting a target R-value and solving for baffle height, but unfortunately it would be difficult to set it up that way. Temperature rating is a pretty imprecise system, so there is not much point in converting R-values directly into temperature ratings. Center-of-baffle height (2*r) is a better proxy for temperature rating, as it is easy to compare to existing quilts.

4. Ellipses

Maybe I haven't sold people on circular baffles yet. You want oval shaped baffles, and I will oblige you. The equations you need are as follows:

In this instance, A is half the length of the ellipse, and B is half the height (you can think of them as r1 and r2). The ratio of A to B is what determines the stability of the shape and the down within it, so you want this ratio to be as close to 1 as possible while still managing the baffle spacing (Code:BoxBaffleShellWt[ShellWt_,A_,B_,x_] :=4*A*EllipticE[ArcSin[x/A],(A^2-B^2)/A^2]*(ShellWt/1296) BoxBaffleWallWt[BaffleWt_,A_,B_,x_]:=2*B*Sqrt[A^2-x^2]/A*(BaffleWt/1296) BoxBaffleFillWt[FillPower_,A_,B_,x_]:=2*NIntegrate[B*Sqrt[A^2-y^2]/A,{y,-x,x}]/FillPower BoxBaffleR[A_,B_,x_]:=(3.93*2*x)/NIntegrate[A/(B*Sqrt[A^2-y^2]),{y,-x,x}] BoxBaffleRatio[ShellWt_,BaffleWt_,FillPower_,A_,B_,x_] := BoxBaffleR[A,B,x]/((BoxBaffleShellWt[ShellWt,A,B,x]+BoxBaffleWallWt[BaffleWt,A,B,x]+BoxBaffleFillWt[FillPower,A,B,x])/(2*x)) FindOptimumBaffle[ShellWt_,BaffleWt_,FillPower_,A_,B_] := NMaximize[BoxBaffleRatio[ShellWt,BaffleWt,FillPower,A,B,x],{x}∈Interval[{0,A}]]x) you want.

A useful way for envisioning some of the tradoffs of increasingly long ovular baffles comes from looking at the problem another way: choosing a fixed baffle width (x), and varying the length of the ellipse before truncation (A) to see how it affects the warmth to weight ratio. For example, take a fairly typical 1.5 inch thick, 3 inch wide baffle. By changing the width of the base shape (thus increasing the height of the baffle walls, and making the shape more rectangular), we increase performance. WhenAis equal tox, you have a sewn-through baffle, and performance is low. AsAtends toward infinity, the baffle shape becomes a rectangle, and thus increasingly unlikely to effectively hold the down in place. The ideal is somewhere in between, though where exactly that is depends on your priorities. The code and the graph it produces are below:

Screenshot_2019-06-28 (unnamed) - Wolfram Development Platform.pngCode:Plot[BoxBaffleRatio[0.885, 0.67, 750, A, 0.75,1.5],{A,1.5,6}, PlotRange->Full]

And that finishes Part II of the series. Part III will be on differential baffles and underquilts, and I hope to put it out in the next week or two. Before that, I might post a revision of Part I, since my (continuing) research and input from the community has left me with a better understanding of the topic, and I believe I will be able to better relay the information in a form that people will be able to use in their own projects.

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