Advertisement I got this beautiful girl in the mail today and had her tank ready to go. Online the description told me it was 2” 1/2. I think I underestimated the tarantulas size though. Tank dimensions: L: 8” W: 5” H: 6 1/2” I know they like generally smaller enclosures, but I thought I’d just make sure I chose the right tank size. Too small?

Let's say that she can't arrange a meeting there but as said above, anyway, can work. But really, honestly, I would offer her something maybe just a bit bigger. Not that much, but that much slightly bigger, yes

Thanks for the insight! I think after her next molt, I'll move her to a bigger enclosure. Way ahead of ya! She has 3 inches to work with

oh how misfortunately said. anyways there is a pretty general formula for this: (L*W) - (3*(dls^2)) >= 0. if returned false then an upgrade needed, the coefficient 3 can be adjusted I'd recommend no less than 2. this works better for terrestrial Ts as its all based on the difference of surface area of enclosure and dls square size and does not include any rules about vertical space. for vert space no greater than 1.5(dls) this is only for terrestrials. fossorials and arboreals have different rules.

I find these very small enclosures kind of sad. For me, this would definitely be too small. A. chalcodes is a moderately active tarantula after all. What happened to the 2x DLS to 3x DLS rule, that @Tenebrarius put into a slightly more complicated form above? This enclosure is no more than 1x DLS wide and 2x DLS long. If the tarantula really stretches it looks as if she can touch both sides at once. Unfortunately there is no science to this and tarantulas survive in really small enclosures but: I would never keep a tarantula in an enclosure this small.

Thanks for providing a formula for me to work with. I suppose I'll get her a bigger enclosure to live in. However the formula is a little confusing. Would you mind explaining it in a simpler way? When you put it that way, it does sound a little depressing. Time to get a bigger enclosure!

it seemed pretty simple, take the surface area if the enclosure and stop it in a variable A; now take the dls and square it, then take the square and multiply it by some constant, which can be variable 2 or higher (recommend 3), now take the product and store it in some variable B; now subtract B from A (A - B) and there will be your measure it will be in inches so if your use the superior metric so you adjust the constant to a better number for centimeters and our dls and surface area as centimeter numbers simple fix. so the answer remaining should be equal to to or more if not return false and get better inputs.

... you call that simple? I haven't got a clue what you just said I'll stick to grammar and spelling lol

We can't all be grommaticly inclined, . this is why I like math, the only commas are optional or in ordered pairs...also optional based on what notation you use. f(1) = 3 where f(x) = y is the same thing as (3, 1). oh simplicity I cant wait for someone to optimize english. The point was to expand using english to explain to OP maybe I did a bad job.

I think you probably did a fine job of explaining it... I am just *really* terrible at calculations lol It seems to be that way though - an inclination towards words or numbers, and struggling a bit (or in my case, a lot!) with the other.

I actually understand you better when you talk math than when you talk English... Ah, sorry, I meant I understand @Tenebrarius better!