Query on the geometry of peeling carrots
December 4, 2018 10:56 AM Subscribe
When I peel carrots, I've noticed it almost invariantly takes nine swipes of the peeler to completely peel the carrot. Counter-intuitively, this number doesn't change according to the size (diameter) of the carrot. Is this some known constant property of the geometry of circles and cylinders and how they intersect with planes?
I invite people to peel a carrot and test if it just my peeler, perhaps.
I invite people to peel a carrot and test if it just my peeler, perhaps.
This can't be true of any size carrot: Imagine a theoretical carrot with, say, a 10 foot diameter. It would take many more swipes. Now consider a carrot with a diameter of a few millimeters: It would take at most three swipes, unless you could somehow make each peel incredibly thin.
My guess is that your carrots do not vary enough in size for you to see variance in the number of swipes needed.
posted by Mr.Know-it-some at 11:14 AM on December 4, 2018 [10 favorites]
My guess is that your carrots do not vary enough in size for you to see variance in the number of swipes needed.
posted by Mr.Know-it-some at 11:14 AM on December 4, 2018 [10 favorites]
Response by poster: My peeler is similar to yours, Grither.
The ten foot example makes intuitive sense but the mm one doesn't, consider the intersection of the plane with the circumference - three cuts would produce all peel no carrot.
Maybe this has more to do with the depth of the cut peel.
posted by Rumple at 11:20 AM on December 4, 2018
The ten foot example makes intuitive sense but the mm one doesn't, consider the intersection of the plane with the circumference - three cuts would produce all peel no carrot.
Maybe this has more to do with the depth of the cut peel.
posted by Rumple at 11:20 AM on December 4, 2018
Perhaps when a carrot has more girth you subconsciously press down on the peeler more because the carrot appears "tougher", thereby taking a deeper slice off the carrot, allowing you to peel it in fewer swipes than you would if you applied normal pressure? Or vice-versa when a thinner carrot appears more delicate.
Try to be consistent when you peel different thicknesses of carrot and see how that affects your results.
posted by bondcliff at 11:22 AM on December 4, 2018 [3 favorites]
Try to be consistent when you peel different thicknesses of carrot and see how that affects your results.
posted by bondcliff at 11:22 AM on December 4, 2018 [3 favorites]
I am here only to say that I have the same experience. 9 swipes on a big carrot, nine swipes on a thinner carrot. Happy to learn that I am not the only one.
posted by sheldman at 11:44 AM on December 4, 2018 [3 favorites]
posted by sheldman at 11:44 AM on December 4, 2018 [3 favorites]
I think even on the giant carrot, you'd be using an appropriate-sized peeler and the giant peeling the giant carrot would need the same number of swipes.
posted by If only I had a penguin... at 11:51 AM on December 4, 2018 [5 favorites]
posted by If only I had a penguin... at 11:51 AM on December 4, 2018 [5 favorites]
If you peel a carrot with three swipes, you end up with a triangle. If you remove nine sections from a circle, you end up with a nonagon.
posted by soelo at 11:52 AM on December 4, 2018 [3 favorites]
posted by soelo at 11:52 AM on December 4, 2018 [3 favorites]
Yeah, I'd hazard this is maybe more of a psychology question than just a geometry one.
It's the intersection of two constraints -- first,your own internal sense of how much loss is acceptable through peeling. Like, you could obviously peel a carrot using a knife with say four strokes, yielding a square prism, but picture how much outer carrot flesh you'd lose that way; it would feel wasteful and wouldn't yield the nice round shape you're (i assume) aiming for.
And second - how many strokes feels like "too many to be worth it." There's some balance point between those two constraints (wanting to save usable food vs wanting to save labor). It would be interesting to find someone who consistently has a different standard number of strokes.
posted by LobsterMitten at 11:54 AM on December 4, 2018 [16 favorites]
It's the intersection of two constraints -- first,your own internal sense of how much loss is acceptable through peeling. Like, you could obviously peel a carrot using a knife with say four strokes, yielding a square prism, but picture how much outer carrot flesh you'd lose that way; it would feel wasteful and wouldn't yield the nice round shape you're (i assume) aiming for.
And second - how many strokes feels like "too many to be worth it." There's some balance point between those two constraints (wanting to save usable food vs wanting to save labor). It would be interesting to find someone who consistently has a different standard number of strokes.
posted by LobsterMitten at 11:54 AM on December 4, 2018 [16 favorites]
So, I do this thing where I count stuff. Steps, the # of times the indicator blinker goes when I switch the turn signal on, the # of swipes it takes to shave my leg, the # of peels for a carrot. I remember from Thanksgiving counting anywhere from 6 to 12 peels depending on carrot size.
Anecdote of one, there you have it.
posted by cooker girl at 12:03 PM on December 4, 2018 [13 favorites]
Anecdote of one, there you have it.
posted by cooker girl at 12:03 PM on December 4, 2018 [13 favorites]
This thread reminded me that I needed to peel a carrot as I like to eat a carrot raw every day.
The carrot I just peeled was closer to two centimetres in diameter than one inch. It took me thirteen swipes to do the circumference plus several more swipes at the bottom, narrower end to get bits that remained unpeeled but were much less than the full length of the carrot.
I used a regular old fashioned peeler, not a paring knife. I suspect that the greater number of swipes has to do with trying to take off the least depth as possible as I only want to take the bitter part of the peel off and as little as possible below it.
*crunch-crunch-crunch *
posted by Jane the Brown at 12:39 PM on December 4, 2018 [3 favorites]
The carrot I just peeled was closer to two centimetres in diameter than one inch. It took me thirteen swipes to do the circumference plus several more swipes at the bottom, narrower end to get bits that remained unpeeled but were much less than the full length of the carrot.
I used a regular old fashioned peeler, not a paring knife. I suspect that the greater number of swipes has to do with trying to take off the least depth as possible as I only want to take the bitter part of the peel off and as little as possible below it.
*crunch-crunch-crunch *
posted by Jane the Brown at 12:39 PM on December 4, 2018 [3 favorites]
Another possibility is that, if you peel a lot of carrots, you may have unconsciously trained your body mechanics to nine swipes because that is most often the number required for satisfactory peeling of the widest range of common carrot diameters.
That is, you may have settled into a nine-swipe rhythm because it is easier/more efficient to repeat that than to constantly assess and reconfigure the mechanical motion. If you want to get quickly through a boring and repetitive task, a replicable rhythm is your best bet.
I suspect if you repeated this investigation with potatoes, you'd find more variability, because potatoes can vary more in size. I think you'd still see the average stroke count converging on a given number the more you peel.
posted by helpthebear at 1:19 PM on December 4, 2018 [4 favorites]
That is, you may have settled into a nine-swipe rhythm because it is easier/more efficient to repeat that than to constantly assess and reconfigure the mechanical motion. If you want to get quickly through a boring and repetitive task, a replicable rhythm is your best bet.
I suspect if you repeated this investigation with potatoes, you'd find more variability, because potatoes can vary more in size. I think you'd still see the average stroke count converging on a given number the more you peel.
posted by helpthebear at 1:19 PM on December 4, 2018 [4 favorites]
Best answer: Hypothesis:
The larger a carrot, the less flexible it is.
The less flexible it is, the harder you can push with the peeler without bending it, and the more the peeler can "bite" into it instead of pushing it out of the way.
And so peeler pressure and carrot rigidity are varying together in a way that maintains the nine-swipe pattern for a reasonably wide range of carrot sizes.
posted by nebulawindphone at 1:23 PM on December 4, 2018 [3 favorites]
The larger a carrot, the less flexible it is.
The less flexible it is, the harder you can push with the peeler without bending it, and the more the peeler can "bite" into it instead of pushing it out of the way.
And so peeler pressure and carrot rigidity are varying together in a way that maintains the nine-swipe pattern for a reasonably wide range of carrot sizes.
posted by nebulawindphone at 1:23 PM on December 4, 2018 [3 favorites]
I never peel my carrots but right now I'm studying for a final exam and so of course this was the perfect time to go dig the peeler out of the drawer and go to town on a carrot ... and I'll be damned, it took me 9 swipes! I even divided it up into the thicker top half and the thinner bottom half, and on both halves I ended up with the same number of peels. My peeler is one of those y-style ones, for what it's worth.
Flagged as fantastic.
posted by DingoMutt at 1:25 PM on December 4, 2018 [3 favorites]
Flagged as fantastic.
posted by DingoMutt at 1:25 PM on December 4, 2018 [3 favorites]
Best answer: I'm at the office and don't have any carrots right now, so I'm just going to imagine a perfectly cylindrical vegetable and peel it using mathematics.
Suppose we had a carrot of radius r and a peeler with cutting depth h. Since scaling this scenario up or down doesn't really change it, I'm going to let r equal 1 (and let h be the ratio of peel-depth to carrot radius). Presumably there is some number n such that n slices peel this cylindrical carrot into a regular n-gonal prism. Cross-sectionally speaking, we're finding a regular n-gon that fits inside a circle of radius 1.
The apothem of a regular polygon is the distance from its center to the midpoint of any side. In order for us to be able to peel a regular polygon our of a circle, we require its apothem to have length at least 1–h; otherwise, one peeling wouldn't take off enough depth. The apothem can be trigonometrically determined to have length cos(π/n), so the formula we need n to satisfy is cos(π/n)>1–h. This can be rearranged to be n>π/arccos(1–h).
In particular, n will be 9 when h is between 0.060 and 0.049. To judge the validity of that claim, we'll need data on actual carrots and peelers. USDA standards indicate that carrots will range in radius from 0.375 to 0.75 inches. There is no such convenient standardization for slicers, but a lot of mandolines can slice as thin as 1.5mm (0.059in), so let's go with something a little smaller, say, 1mm. Then h will be between 0.10 and 0.21. That's way too big; a slicer with that depth would peel a carrot cleanly in between 5 and 7 strokes. So maybe slicers cut a lot finer than I think. Or we're being very inefficient and overlapping our peel strips by a lot.
posted by jackbishop at 2:12 PM on December 4, 2018 [6 favorites]
Suppose we had a carrot of radius r and a peeler with cutting depth h. Since scaling this scenario up or down doesn't really change it, I'm going to let r equal 1 (and let h be the ratio of peel-depth to carrot radius). Presumably there is some number n such that n slices peel this cylindrical carrot into a regular n-gonal prism. Cross-sectionally speaking, we're finding a regular n-gon that fits inside a circle of radius 1.
The apothem of a regular polygon is the distance from its center to the midpoint of any side. In order for us to be able to peel a regular polygon our of a circle, we require its apothem to have length at least 1–h; otherwise, one peeling wouldn't take off enough depth. The apothem can be trigonometrically determined to have length cos(π/n), so the formula we need n to satisfy is cos(π/n)>1–h. This can be rearranged to be n>π/arccos(1–h).
In particular, n will be 9 when h is between 0.060 and 0.049. To judge the validity of that claim, we'll need data on actual carrots and peelers. USDA standards indicate that carrots will range in radius from 0.375 to 0.75 inches. There is no such convenient standardization for slicers, but a lot of mandolines can slice as thin as 1.5mm (0.059in), so let's go with something a little smaller, say, 1mm. Then h will be between 0.10 and 0.21. That's way too big; a slicer with that depth would peel a carrot cleanly in between 5 and 7 strokes. So maybe slicers cut a lot finer than I think. Or we're being very inefficient and overlapping our peel strips by a lot.
posted by jackbishop at 2:12 PM on December 4, 2018 [6 favorites]
I imagine we are overlapping peel strips, because it's hard to peel with precision, and erring on the side of overlapping is better than erring on the side of leaving peel behind.
posted by nebulawindphone at 2:16 PM on December 4, 2018 [2 favorites]
posted by nebulawindphone at 2:16 PM on December 4, 2018 [2 favorites]
Best answer: If you are always peeling with nine cuts, regardless of diameter, it means that width of your cut is always approximately 40 degrees. But if the angle is constant, then the depth of cut is proportional to the diameter. If the carrot is twice as big in diameter, you must be making your cut twice as deep.
So what seems to be happening is that you are unconsciously making a deeper cut for a bigger carrot so that in appearance, your cut is approximately a constant 40 degrees wide. In other words, when you look down on the carrot, the width of your cut is a constant percentage of the width of the carrot.
If you work out the numbers assuming 40 degrees (nine cuts), then if you look straight down on the carrot diameter, the width of your cut is almost exactly one-third of the diameter seen by your eye. So you see one-third round on the left, one-third round on the right and centered exactly between them is the one-third that is cut flat by your cut. You achieve this symmetric proportion by unconsciously adjusting the depth of the cut.
Seems to be kind of a natural symmetry for the eye which causes you to adjust the depth of the cut according to the diameter into thirds. The result is nine cuts.
posted by JackFlash at 3:20 PM on December 4, 2018 [8 favorites]
So what seems to be happening is that you are unconsciously making a deeper cut for a bigger carrot so that in appearance, your cut is approximately a constant 40 degrees wide. In other words, when you look down on the carrot, the width of your cut is a constant percentage of the width of the carrot.
If you work out the numbers assuming 40 degrees (nine cuts), then if you look straight down on the carrot diameter, the width of your cut is almost exactly one-third of the diameter seen by your eye. So you see one-third round on the left, one-third round on the right and centered exactly between them is the one-third that is cut flat by your cut. You achieve this symmetric proportion by unconsciously adjusting the depth of the cut.
Seems to be kind of a natural symmetry for the eye which causes you to adjust the depth of the cut according to the diameter into thirds. The result is nine cuts.
posted by JackFlash at 3:20 PM on December 4, 2018 [8 favorites]
Yes, it’s an interaction of depth and angle, together with the fact that you’ve never peeled a meter-wide carrot, nor a milimeter-wide carrot.
Me, I wonder who peels carrots and why. I feel like life is too short to peel healthy carrots or potatoes or lots of other veggies, YMMV.
posted by SaltySalticid at 4:05 PM on December 4, 2018 [1 favorite]
Me, I wonder who peels carrots and why. I feel like life is too short to peel healthy carrots or potatoes or lots of other veggies, YMMV.
posted by SaltySalticid at 4:05 PM on December 4, 2018 [1 favorite]
Best answer: I have some similar thoughts to jackbishop above, and will borrow their model and notation, but let the radius r vary while keeping the peeling depth h (a.k.a. the sagitta) fixed. I'll also use θ for the angle pi/n.
In this case we have h = r(1–cos θ) ≈ r·θ2/2 (the approximation is from Taylor series), so θ ≈ √(2h/r). This implies that n varies as roughly proportional to √r. So if you double the radius of the carrot while cutting to the same depth, you do need more cuts, but only about √2 ≈ 1.4 times as many cuts, not twice as many as one might expect. (And this actually makes sense because bigger circles are less curved.) That might be part of what's going on.
(As long as I'm typing this comment, a non-root-vegetable application of roughly the same math described here is that as you gain elevation above the earth, the distance you can see to the horizon grows roughly as the square root of your altitude, which is kind of cool to know. From eye level at 5 feet, you can see about 3 miles on flat ground; from 500 feet up, you can see about 30 miles. So that's kind of nifty.)
posted by aws17576 at 7:31 PM on December 4, 2018 [2 favorites]
In this case we have h = r(1–cos θ) ≈ r·θ2/2 (the approximation is from Taylor series), so θ ≈ √(2h/r). This implies that n varies as roughly proportional to √r. So if you double the radius of the carrot while cutting to the same depth, you do need more cuts, but only about √2 ≈ 1.4 times as many cuts, not twice as many as one might expect. (And this actually makes sense because bigger circles are less curved.) That might be part of what's going on.
(As long as I'm typing this comment, a non-root-vegetable application of roughly the same math described here is that as you gain elevation above the earth, the distance you can see to the horizon grows roughly as the square root of your altitude, which is kind of cool to know. From eye level at 5 feet, you can see about 3 miles on flat ground; from 500 feet up, you can see about 30 miles. So that's kind of nifty.)
posted by aws17576 at 7:31 PM on December 4, 2018 [2 favorites]
Response by poster: So if I collect the 9 peelings from a small carrot of radius n and a large carrot of radius 2n, then I should find that the average peel of the larger carrot is 1.4 times that of the smaller carrot?
I have calipers but no carrots so this will be an interesting test in the next day or two.
posted by Rumple at 7:58 PM on December 4, 2018
I have calipers but no carrots so this will be an interesting test in the next day or two.
posted by Rumple at 7:58 PM on December 4, 2018
Best answer: No, the peel is always directly proportional to the diameter (or radius) of the carrot. If the carrot is twice the radius, the peel must be twice a wide.
As I showed in my analysis above, if it is true that you peel the carrot with nine cuts, no matter the size, then the width of every peel will be very close to one-third the width (diameter) of the carrot.
Not only that, if you maintain the same nine cuts down the length of the carrot, you should find that the width of the peel decreases as you go toward the tapered end so that the peel is always about one-third of the width of the carrot at that point. To do so, you would be shallowing the depth of your cut as you go toward the small end.
I'm guessing that visually, you adjust the depth of your cut so that you maintain a one-third orange strip on the left, a one-third orange strip on the right and a one-third flat strip for your cut. Visually, it is easy to regulate the cutting depth down the length of the carrot.
Through trigonometry you can show if you maintain a one-third width of the peel, that the flat cut (technically the chord of a circle) is contained by a 40 degree angle. There are nine of these 40 degree angles in the full circle around the carrot.
I can show the trigonometry if you would like details.
posted by JackFlash at 8:27 PM on December 4, 2018 [1 favorite]
As I showed in my analysis above, if it is true that you peel the carrot with nine cuts, no matter the size, then the width of every peel will be very close to one-third the width (diameter) of the carrot.
Not only that, if you maintain the same nine cuts down the length of the carrot, you should find that the width of the peel decreases as you go toward the tapered end so that the peel is always about one-third of the width of the carrot at that point. To do so, you would be shallowing the depth of your cut as you go toward the small end.
I'm guessing that visually, you adjust the depth of your cut so that you maintain a one-third orange strip on the left, a one-third orange strip on the right and a one-third flat strip for your cut. Visually, it is easy to regulate the cutting depth down the length of the carrot.
Through trigonometry you can show if you maintain a one-third width of the peel, that the flat cut (technically the chord of a circle) is contained by a 40 degree angle. There are nine of these 40 degree angles in the full circle around the carrot.
I can show the trigonometry if you would like details.
posted by JackFlash at 8:27 PM on December 4, 2018 [1 favorite]
Best answer: So if I collect the 9 peelings from a small carrot of radius n and a large carrot of radius 2n, then I should find that the average peel of the larger carrot is 1.4 times that of the smaller carrot?
No, the peelings from the large carrot would be twice as thick. What I was saying was that it would take about 1.4*9 ≈ 13 peelings from the larger carrot to make those peelings the same thickness as 9 from the smaller carrot... though now I see that that doesn't really apply to your question. Somewhere in the course of writing my answer I forgot that you were looking for an explanation for "always exactly nine", and gave a possible explanation for "usually surprisingly close to nine", which really is not the same. My bad!
posted by aws17576 at 11:18 PM on December 4, 2018 [1 favorite]
No, the peelings from the large carrot would be twice as thick. What I was saying was that it would take about 1.4*9 ≈ 13 peelings from the larger carrot to make those peelings the same thickness as 9 from the smaller carrot... though now I see that that doesn't really apply to your question. Somewhere in the course of writing my answer I forgot that you were looking for an explanation for "always exactly nine", and gave a possible explanation for "usually surprisingly close to nine", which really is not the same. My bad!
posted by aws17576 at 11:18 PM on December 4, 2018 [1 favorite]
Me, I wonder who peels carrots and why. I feel like life is too short to peel healthy carrots or potatoes or lots of other veggies, YMMV.
posted by SaltySalticid
People who have more taste receptors for bitterness. People who are avoiding roughage at all costs because of Crohn's or other digestive issues. People who enjoy peeling carrots. People whose carrots come in a natural state with dirt and bumps and knobs on them, and find peeling more satisfying than scrubbing with a vegetable brush.
What _I_ wonder is if anyone scrapes carrots anymore, and if so how many scrapes on the average do they take? Is it nine, the same as peeling? Or when you scrape a carrot is the swipe more shallow than when you peel so there are necessarily more? Or do you only scrape the rough looking bits so you skip sections?
posted by Jane the Brown at 1:35 PM on December 5, 2018 [1 favorite]
posted by SaltySalticid
People who have more taste receptors for bitterness. People who are avoiding roughage at all costs because of Crohn's or other digestive issues. People who enjoy peeling carrots. People whose carrots come in a natural state with dirt and bumps and knobs on them, and find peeling more satisfying than scrubbing with a vegetable brush.
What _I_ wonder is if anyone scrapes carrots anymore, and if so how many scrapes on the average do they take? Is it nine, the same as peeling? Or when you scrape a carrot is the swipe more shallow than when you peel so there are necessarily more? Or do you only scrape the rough looking bits so you skip sections?
posted by Jane the Brown at 1:35 PM on December 5, 2018 [1 favorite]
Response by poster: OMG scraped carrots are nasty with those little balls of peel which get everywhere.
Special thanks to JackFlash, jackbishop, and aws17576 for their answers. I was hoping for a purely geometric solution and in my mind's eye I can see it but I guess that is just truthiness geometry. It is interesting though that the thickness seems to be the variable, and that is cued perhaps by a semi-intentional rule of thirds as one is peeling carrots, and or a kind of haptic feedback from the size of carrot controlling the pressure put on the blade.
It seems obvious now if I peeled a large zucchini it would take more than 9 swipes, but then, if I had a custom zucchini peeler, perhaps not.
posted by Rumple at 1:56 PM on December 5, 2018
Special thanks to JackFlash, jackbishop, and aws17576 for their answers. I was hoping for a purely geometric solution and in my mind's eye I can see it but I guess that is just truthiness geometry. It is interesting though that the thickness seems to be the variable, and that is cued perhaps by a semi-intentional rule of thirds as one is peeling carrots, and or a kind of haptic feedback from the size of carrot controlling the pressure put on the blade.
It seems obvious now if I peeled a large zucchini it would take more than 9 swipes, but then, if I had a custom zucchini peeler, perhaps not.
posted by Rumple at 1:56 PM on December 5, 2018
Best answer: Well, this isn't totally it, but geometrically speaking, one way you can approximate a circle is via polygons with more and more sides. So you start with a triangle (3-gon), then a square (4-gon), pentagon, etc.
You can see the progression as you go from a 3-gon to a 10-gon in this graphic.
The Big Picture geometrical point is that triangles, squares, and even pentagons are pretty BAD approximations of a circle. Just looking at that graphic, the first ones on the list that even look approximately like a circle are the ones with about 7, 8, or 9 sides.
So when you peel a carrot, you are trying to approximate a circle (decent characterization of a typical carrot cross-section) with a series of straight lines (decent characterization of what the typical straight peeler does to a carrot).
No matter how large or how small the carrot, the least amount of sides you can POSSIBLY get away with, and have the end result but anywhere at all close to circle-ish, is somewhere in the range of 7-8-9.
So, even if your carrot were say 0.001 mm in diameter, or 0.1 mm, or 1mm, or 3mm, say--all very, very small scales from a human perspective--you still couldn't peel it with just 3 or 4 swipes because the end result would be very much triangular or square and not even close to round.
Point is, 7-9 sides is about the minimum needed to approximate roundness to any degree whatsoever, and this is pretty true irregardless of scale. It's true whether your carrot is 0.1 mm in diameter or 100 meters.
But it's particularly true at the small end of the scale. Let's say you're satisfied with the relative "roundness" of 10-meter diameter object with 100 sides, and so a 5-meter diameter takes 50 sides, 2.5-meter takes 25 sides, and so on down. But at some point you reach about 9 sides and you can't really go any smaller. Like if your 0.9 meter diameter carrot is satisfactory with 9 sides, you don't say, "Well then a 0.5 meter carrot only needs 5 sides and a 0.2 meter carrot only needs 2 sides!"
It just doesn't keep scaling down linearly that way. You still need approximately 9 sides to achieve an approximation of roundness no matter how small in scale you go.
So if you like technical terms, one way you could characterize this is something like a scale invariant--maybe not over the whole scale of sizes, but at least at the lower end.
Another way you could think of it is as an example of diminishing returns. As you move from from 2 to 3-4-5-6 swipes to peel your carrot, you're doing some more work at each step, but you're saving a l-o-t of carrot. But once you're up to 8 or 9 swipes, that's kind of a sweet spot of peeling-work vs carrot-savings. Like if you went from 9 to 18 swipes it would be twice the work and the end result wouldn't be very noticeably different at all.
This gets back to the first point I made, which is that moving from say 3 to 9 sided polygons gets you a really major increase in the roundness of the result. Whereas moving from 9 to 18 gains relatively little more.
If you want to quantify the relative diminishing returns, you could mess around with online 'pi calculators' like this or this. Summarizing what you find there (in terms of difference in circumference length):
- Triangle - about 20% off vs circle
- 9-gon - about 2% off
- 18-gon - about 0.5% off
- 36-gon - about 0.1% off
So . . . diminishing returns.
posted by flug at 8:39 PM on December 5, 2018 [2 favorites]
You can see the progression as you go from a 3-gon to a 10-gon in this graphic.
The Big Picture geometrical point is that triangles, squares, and even pentagons are pretty BAD approximations of a circle. Just looking at that graphic, the first ones on the list that even look approximately like a circle are the ones with about 7, 8, or 9 sides.
So when you peel a carrot, you are trying to approximate a circle (decent characterization of a typical carrot cross-section) with a series of straight lines (decent characterization of what the typical straight peeler does to a carrot).
No matter how large or how small the carrot, the least amount of sides you can POSSIBLY get away with, and have the end result but anywhere at all close to circle-ish, is somewhere in the range of 7-8-9.
So, even if your carrot were say 0.001 mm in diameter, or 0.1 mm, or 1mm, or 3mm, say--all very, very small scales from a human perspective--you still couldn't peel it with just 3 or 4 swipes because the end result would be very much triangular or square and not even close to round.
Point is, 7-9 sides is about the minimum needed to approximate roundness to any degree whatsoever, and this is pretty true irregardless of scale. It's true whether your carrot is 0.1 mm in diameter or 100 meters.
But it's particularly true at the small end of the scale. Let's say you're satisfied with the relative "roundness" of 10-meter diameter object with 100 sides, and so a 5-meter diameter takes 50 sides, 2.5-meter takes 25 sides, and so on down. But at some point you reach about 9 sides and you can't really go any smaller. Like if your 0.9 meter diameter carrot is satisfactory with 9 sides, you don't say, "Well then a 0.5 meter carrot only needs 5 sides and a 0.2 meter carrot only needs 2 sides!"
It just doesn't keep scaling down linearly that way. You still need approximately 9 sides to achieve an approximation of roundness no matter how small in scale you go.
So if you like technical terms, one way you could characterize this is something like a scale invariant--maybe not over the whole scale of sizes, but at least at the lower end.
Another way you could think of it is as an example of diminishing returns. As you move from from 2 to 3-4-5-6 swipes to peel your carrot, you're doing some more work at each step, but you're saving a l-o-t of carrot. But once you're up to 8 or 9 swipes, that's kind of a sweet spot of peeling-work vs carrot-savings. Like if you went from 9 to 18 swipes it would be twice the work and the end result wouldn't be very noticeably different at all.
This gets back to the first point I made, which is that moving from say 3 to 9 sided polygons gets you a really major increase in the roundness of the result. Whereas moving from 9 to 18 gains relatively little more.
If you want to quantify the relative diminishing returns, you could mess around with online 'pi calculators' like this or this. Summarizing what you find there (in terms of difference in circumference length):
- Triangle - about 20% off vs circle
- 9-gon - about 2% off
- 18-gon - about 0.5% off
- 36-gon - about 0.1% off
So . . . diminishing returns.
posted by flug at 8:39 PM on December 5, 2018 [2 favorites]
This thread is closed to new comments.
posted by Grither at 10:59 AM on December 4, 2018 [1 favorite]