Friday, February 22, 2013

Unfinished business: various

Occasionally, when I'm trying to find an old blog post, I run into other things along the way that aren't what I was looking for but make me think oh, I should say something about how that turned out. But then I don't do that, because the idea only occurs when I'm in the middle of writing something else. But today is the day, because I don't have any great ideas at the moment for writing anything else. So, here are some follow-ups on the posts from February, January, and December:

Random plant event: Hatiora NOID (15 February 2013):

The buds are still there, though they're not growing very fast, there aren't any new ones, and the ones there are haven't opened yet.

Random plant event: Anthurium #58 ("Betty Larceny") (7 February 2013):

Betty was just messing with me. I am not 100% certain, but it looks like the pink thing I thought was maybe a flower was actually just a cataphyll that was more strongly colored than the plant normally produces. Oh well. It kinda did seem too good to be true.

Random plant event: Spathiphyllum (20 January 2013):

The Spathiphyllum seedlings are beginning to look like Spathiphyllums, a bit. Slightly bigger leaves, and more leaves. Even getting some of the sunken primary veins going.

Of the 18 pots, 16 survive. (The other two got too dry. My bad.)

As with children, it turns out that Spathiphyllums are a lot more interesting, likeable, and adorable when you've made them yourself. I'm checking up on them all the time, I get excited when I see that they're growing, I'm trying to cross the parents so I can have even more, etc.

Unfinished business: Coffea arabica (28 December 2012):

It's been eight weeks since I took the Coffea seeds out of their fruits and soaked them in water, so I have planted the eighteen seeds from that batch in vermiculite and am trying to germinate them in the basement, as of Monday night. A second set of nine seeds soaked in water for one day, dried for five days, and then soaked in water for one day again. I'm just checking to see if drying time really makes that much difference.

Nothing so far, but I wasn't really expecting anything to happen yet anyway. I'm not sure I should be expecting anything to happen ever, since I'm not sure I'm doing any of this correctly.

Random plant event: Gasteria bicolor (18 December 2012):

The Gasteria seedlings haven't rotted out yet, which is exciting all on its own. They're also up to two or three leaves apiece.

There is one item of good/bad/good/bad/good news, however: the Gasteria seedlings are still alive, and have grown more leaves, which is good. But the container is overrun by springtails (visible in the image as gray specks on the surface of the leaves), which is bad. Though the Gasterias have not already been devoured by the springtails, which is probably a good sign that they won't be. Then again, this is more springtails in one place than I've ever seen in my life, so: bad. But it's at least let me get the very largest and clearest picture to date of a springtail, which is somewhat good.

List: Missing From Retail, Part 4 of 5 (14 December 2012):

The Osmanthus fragrans got moved to a cool, bright location (in the plant room), but it was too little too late, apparently, since it completely defoliated. And then the stems dried up and became brittle enough to snap off. And then the trunk went too. It seemed pretty dead, so I threw it out.

What with the spider mites (when just after it first arrived), the lack of new flowers (all last summer), and not having a suitable place to overwinter it, I'm thinking Osmanthus is just not something I'm able to grow at the moment.

List: Missing From Retail, Part 3 of 5 (8 December 2012):

Haemanthus albiflos, on the other hand, has been doing okay for me. It dropped some leaves when it first arrived, which concerned me, but it's apparently settled in well enough, because it's produced one new leaf for me (furthest right in the photo).

It's not amazing or anything, but it's a bit of a relief. I'd been a little worried about it. It wasn't the most auspicious time of year to be shipped.

Coming eventually: September to November!

Wednesday, February 20, 2013

Tuesday, February 19, 2013

A Word Problem

Mr. Subjunctive has 38 seeds to start in vermiculite.1 The container he wants to start them in is an undivided rectangle measuring 12 inches by 8 inches.

Question 1. How many rows and columns of seeds should he make if he wants to place the seeds as far apart from one another as possible, in a square grid like that in Figure 1?

Figure 1. Seedling silhouettes are colored various shades of red, green, and brown because I think the diagrams are more interesting if they have multiple colors in them. Also, one parent is red, and one parent is green, so at least in theory, the offspring could wind up anywhere in between the two, which makes it almost realistic or something.

Question 2. How many rows and columns of seeds should he make if he wants to place the seeds as far apart from one another as possible, but in a triangular/hexagonal grid instead, as in Figure 2?

Figure 2

Extra credit. How many of the 38 seeds Mr. Subjunctive plants will actually germinate and grow long enough to flower and produce their own seeds? Explain your answer.

Answers are below.





Question 1. Although we could just plant a single row of 38 seedlings down the center of the rectangle and call it done, the seedlings would have to be very close together in the row, with a lot of wasted space on either side, so they'll crowd one another out. In order to give each seed as much space as possible, we're going to use algebra to try and find a rectangle of seedlings that has the proportions which best fit the container.

Since we know that the container is a rectangle, 12 inches by 8 inches, we know that we're looking for a rectangle of seedlings with proportions similar to 12 x 8.

So. Let x = the number of seedlings in a column, parallel to the short side of the container. The number of seedlings in a row (parallel to the long side of the container) then has to be 1.5x, because the long side of the container is (12/8) = 1.5 times longer than the short side.

To solve for x and find out how many seedlings to plant per column, we multiply x (the number per column) with 1.5x (the number per row) and set that equal to 38 (the total number of seedlings we have):

38 = 1.5x2

Dividing both sides by 1.5:

25.3 = x2

Taking the square root of both sides to solve for x:

x = √(25.3) = 5 (more or less)

(3/2)x = 7.5 (more or less)

So the most evenly-spaced gridded arrangement would be five rows of eight seedlings, with two empty spaces somewhere in the grid, as in Figure 3:

Figure 3. Solution to Question 1.

Question 2. Although the whole triangular-grid thing is unusual and scary, we can look at this as just the sum of two rectangles. See Figure 4.

Figure 4. Here, the red seedlings are arranged as a 4x3 rectangle, while the green seedlings are in a 3x2 rectangle. When looking at the red and green seedlings combined, though, they're still in a triangular grid: we're just looking at them as two separate rectangles, added together, because that makes it possible to calculate what we need to calculate.

The outer rectangle will have x rows and y columns; the inner rectangle will then have (x-1) rows and (y-1) columns because . . . it just kind of has to?2

So, the seedlings in the outer rectangle = xy.

The seedlings in the inner rectangle = (x-1)(y-1) = xy-x-y+1.

The sum of the seedlings in the two rectangles combined should be equal to the number of seedlings we have, so:

38 = (xy) + (xy-x-y+1)
38 = xy+xy-x-y+1
38 = 2xy-x-y+1

Well. We're not going to be able to solve it looking like that. So let's think about what we know of x and y, compared to one another.

If we call the short side y, then we know that x=1.5y, using the same reasoning as in the first question. So, everywhere that x appears in the original sum, we can substitute 1.5y instead, giving us this:

38 = 2(1.5y)y-(1.5y)-y+1
38 = 3y2-1.5y-y+1
38 = 3y2-2.5y+1

Subtracting 38 from both sides:

0 = 3y2-2.5y-37

Which is a quadratic equation, and those are easily solved in the internet age.3 Putting in the numbers a=3, b=-2.5, and c=-37, the formula gives two possible answers for y:

3.953 blah blah blah, and
-3.120 blah blah blah.

As interesting as it might be to try to plant negative three rows of seedlings, or 3.95 rows, for that matter, for our purposes there's only one valid answer, which is that y is basically equal to 4.

If y=4, then x has to be (1.5)(4) = 6.

So the large rectangle is 6 columns by 4 rows --

Figure 6

Making the small rectangle 5 columns by 3 rows --

Figure 7

Which means 24 in the large rectangle and 15 in the small rectangle. 24 plus 15 is 39, and we only have 38 seedlings, so one of the holes will have to be empty, as in figure 8.

Figure 8. Solution to Question 2.

Extra credit.
The answer is zero, because Mr. Subjunctive has never attempted to germinate Cryptbergia seeds before, and couldn't find anything on the internet to tell him how to do it, so he's probably going about this completely wrong. There's a possibility that the seeds were already dead by the time he noticed them, also.

I would also accept: zero, because however many of them germinate and grow, they're just going to get scale, thrips, or fungus and be thrown out anyway.
Special note regarding algebra: all kidding aside, though, algebra is way more useful in day to day life than you might think. Trigonometry is way overrated, and I don't think I've ever needed calculus or differential equations for anything outside of schoolwork (which is lucky, 'cause I no longer remember how to do any of it), and arithmetic is trivial, what with calculators and Google and all,4 but algebra comes up all the time and really is just as useful as your algebra teachers always told you it would be.

1 What kind of seeds? Well, if you must know, I tried to cross the Cryptbergia x rubra with the Billbergia nutans that was blooming at the same time, making them technically some sort of Cryptbergia. Let's call them Cryptbergia x subjunctiva. Here's what the dried inflorescences looked like:

And here is what the seeds looked like, with the dead flower parts still attached:

The kind of seeds doesn't actually matter to the math, though, obviously.
2 I know there's nothing in the whole triangular-grid concept that would prevent us from extending the inner rectangle on one side, like in Figure 5:

Figure 5. Here we've got the same 4x3 rectangle of red seedlings, but now there's a 3x3 rectangle of green ones, instead of 3x2, because the green rectangle is not entirely enclosed by the red one. Instead, one row of green ones is outside the rectangle of red ones.

Which since you're so smart, why don't you just go ahead and figure out how to arrange the seedlings if the inner rectangle is extended on one side then. That oughta teach you a lesson about thinking independently.
3 This is the point where most math teachers would include something about the quadratic equation. While it's true that the quadratic equation can be a handy thing to know sometimes, 1) because of the internet, it's never more than a Google search away, so memorizing it is a lot less valuable than it used to be, and 2) Mr. Subjunctive isn't asking you this question because he cares about whether or not you know the quadratic equation; he just cares about how to arrange the holes he's going to poke in the vermiculite. No need to churn your butter by hand when you can just go to the supermarket.
4 In case you didn't know: you can type arithmetic problems ("2*8+13") directly into Google searches now, and it'll give you the answer ("29"). This is frequently more trouble than just using my trusty TI-81 calculator, so I don't actually Google for arithmetic that often, but still. Nice to know the option is there.

Monday, February 18, 2013

Question for the Hive Mind: Amorphophallus

Actually two questions.

The first: is this tuber coming out of dormancy?

I don't recall the little pink-purple cone being there before, but I also haven't paid a lot of attention to the Amorphophallus konjac since unpotting it last fall. So I'm not sure if this is new, or just something I've failed to notice for several months.

The second question: is this thing on the bottom of the tuber an offset I can divide off, or is it something else?

I ask because when I went searching for pictures of offsets, it looked like several small offsets is the usual way of things: I couldn't find pictures of any single large ones. Also it's sort of weird because I think it shrunk? Or else the parent tuber grew during dormancy. Which doesn't make any sense. Here's what it looked like in October:

Sunday, February 17, 2013

Pretty picture: Oncidium Twinkle 'Sun Light'

Tag: ONC. TWINKLE 'SUN LIGHT' (cheirophorum x sotoanum)