Wyloch's Armory Cardstock Miniature Help

[Undyne]

So, I'm a fan of the Youtuber Wyloch's Armory. I want some help with something though... How would you do a size other than Medium for creatures with his cardstock miniatures?

EDIT: How did this go from a talk on how to do non-medium creatures to arguing over the height of a kobold and the utter stupidity that is the Imperial Measurement system...? And before you ask, I am American, but I agree that most parts of the Imperial system are dumb.

[Destro_Yersul]

Could you provide an example? I'm familiar with Wyloch, but I've only watched his terrain videos.

[Cheesegear]

So, I'm a fan of the Youtuber Wyloch's Armory. I want some help with something though... How would you do a size other than Medium for creatures with his cardstock miniatures?

I know this one.

So, assuming that you're using standard scale (1"=5ft [60"]), which you should be, if you're following normal D&D rules (and you should be, if you're following Wyloch's, 'cause he does), it's pretty simple math; 1/60 scale. 1" scale, equals 60" real.

So, if you know the scale - and you do - all's you really need is a wiki entry to give you the real heights of monsters:

1. Wiki says a Kobold is 2'3". Or, 39" tall. Divide that by 60 (scale), and the model should be 0.65" high.
2. To me - an Australian - that's an asinine measurement. So I multiply inches, into millimeters. 0.65" * 25.4mm
3. I know, that my Kobold - a Small creature - needs to be 16mm high to make ~1m tall creature. (Remembering that a 5'6" (168cm) person would be 28mm model to the eyeline). 16mm, is roughly half of 30mm, so I know my maths checks out. As a Small creature, I put it on a 20mm base (Smalls take up the same 'space' as a Medium, but I like differentiating)

TL:DR
28mm scale models, are at a scale of 1/60 (i.e; 1" = 60"). If you know that, it just takes a bit of maths - and a wiki - and you can make whatever models you want.

Spoiler: Here's some I made last week

Show

[IMG][/IMG]

1. Adult (Huge) Blue Dragon. The model is ~60mm to the eyeline, on a 75mm (3") base.
2. Ancient (Gargantuan) Black Dragon. The model is ~80mm to the eyeline, on a 100mm (4") base.

In case Imgur has a fit.

[Aeson]

Wiki says a Kobold is 2'3". Or, 39" tall. Divide that by 60 (scale), and the model should be 0.65" high.

2' 3" is 27 inches (about 69cm), not 39 inches (3' 3", or about 1m); at a 1:60 scale, the model should be about seven sixteenths or fifteen thirty-seconds of an inch high (half an inch if you're willing to be a bit less precise, maybe three eighths if you're looking at a particularly short kobold).

To me - an Australian - that's an asinine measurement. So I multiply inches, into millimeters. 0.65" * 25.4mm

The Imperial / US Customary system tends to divide inches into power of two fractions (halves, quarters, eighths, sixteenths, thirty-seconds, sixty-fourths, hundred-and-twenty-eighths, etc.) rather than into hundredths, so 0.65 inches would probably be treated as five eighths (0.625) or eleven sixteenths (0.6875) of an inch, maybe as twenty-one thirty-seconds (0.65625) of an inch, depending on how precise you care to make it.

[Cheesegear]

2' 3" is 27 inches (about 69cm), not 39 inches (3' 3", or about 1m)...

...Lucky for me, I haven't made any Kobolds recently, so a typo of 3'3 - instead of 2'3 - doesn't actually matter. The math - despite starting from a faulty premise - checks out. The problem is that it's wrong for a Kobold. For something that was 3'3", it'd be fine - just not a Kobold.

The Imperial / US Customary system tends to divide inches into power of two fractions (halves, quarters, eighths, sixteenths, thirty-seconds, sixty-fourths, hundred-and-twenty-eighths...

That seems real dumb. Asinine, you could say.

so 0.65 inches would probably be treated as five eighths (0.625) or eleven sixteenths (0.6875) of an inch, maybe as twenty-one thirty-seconds (0.65625) of an inch, depending on how precise you care to make it.

I, personally, like that 0.65" would equal 16.51mm (or 1.65cm) every single time.

[Aeson]

I, personally, like that 0.65" would equal 16.51mm (or 1.65cm) every single time.

Except, of course, when you're approximating an inch as 25mm rather than 25.4mm, as you did when specifying the dimensions of your figure bases, 3" and 4" being in reality 76.2mm and 101.6mm respectively rather than the 75mm and 100mm that you listed them as being equivalent to.

Also, this is a question of rounding for convenience - 0.65 inches is 0.65 inches, but if you're going to measure it out or specify a dimension you'll probably round it to five eighths (0.625) or eleven sixteenths (0.6875) of an inch unless you're doing something that calls for greater precision. Similarly, if you were specifying this in the metric system, you'd most likely round to 16.5mm - or perhaps even 16mm or 17mm - rather than specifying 16.51mm, because that level of precision is unnecessary and even - dare I say - asinine for something such as this.

[warty goblin]

That seems real dumb.

It's quite handy in a lot of applications. Woodworking for instance is filled with cases where you need to half something. Half of 7/16 is just 7/32, which simply requires counting on the next smallest set of register lines on the measure.

Imperial is very easy to use for a lot of common tasks. It sucks for multiplying/dividing by ten, but in practice I hardly ever do that. I half or double things all the time.

(Obvious indefensible stupidities: teaspoons being 1/3 of a tablespoon when everything else from gallon on down is a power of two, literally everything to do with miles)

[Cheesegear]

Except, of course, when you're approximating an inch as 25mm rather than 25.4mm, as you did when specifying the dimensions of your figure bases...

I didn't appoximate. I made it 25mm, on purpose. There's a reason for that. Which is simply that I want the bases to be smaller than the space they're actually in. This prevents you from jamming your models together, because the squares are 1" wide, and having your base, also 1" wide, often runs into problems when putting models in adjacent squares. This is why Wyloch - the subject of the thread, after all - puts gaps between his spaces when he makes his tiles. Having your bases, the same size as the square, is bad.

If I wanted 1" bases, I'd have made 1" bases. But I don't, so I didn't. I used to make 1" bases, for 1" squares, because that's the obvious thing to do. Then we ran into problems, and I fixed the problem.

So, my bases are 25mm, to fit into a 25.4mm square, that actually has a practical use. I didn't just do it because measuring is hard.

Point is:
1. Using whatever resources you have available, try and find the 'real' height of the creature you're trying to make. (An Ancient/Gargantuan Black Dragon is 4.9m tall)
2. Divide by 60, and that's the height for your model. (490cm/60 = 8.17cm tall - at the head - model)
3. If making a non-Medium creature, don't forget to change your base size. (Gargantuan would be a 4x4" square)

[Aeson]

I didn't appoximate. I made it 25mm, on purpose. There's a reason for that. Which is simply that I want the bases to be smaller than the space they're actually in.

Your statement:

1. Adult (Huge) Blue Dragon. The model is ~60mm to the eyeline, on a 75mm (3") base.
2. Ancient (Gargantuan) Black Dragon. The model is ~80mm to the eyeline, on a 100mm (4") base.

equates the actual dimensions of the bases, which you provided in metric units, with the nominal dimensions of the bases, which you provided in English / Imperial / US Customary units. Whether you admit it or not, this is an approximation with an implicit assumption of 25mm = 1in and 0.65 inches is therefore 16.25mm rather than the 16.51mm that it "always" is.

I will also point out that while 60 is not a particularly convenient divisor in base-2, it's also not a particularly convenient divisor for base-10. 60 is however fairly close to 64, which is a convenient divisor in base-2 - or when using tools marked in base-2 fractions, as is often the case when working with Imperial / US Customary units - and for something like the kobold you're not going to be too far off if you approximate x/60 as x/64. Big things would be very slightly more inconvenient to handle, but not by much - the Ancient Black Dragon you mentioned, for example, is about 193" tall. 193 is close to 180, which is 60*3, so at 1:60 scale the Ancient Black Dragon is 3 ¹³/₆₀ inches (3.2167" or 81.7mm) tall - which I can approximate as 3 ¹³/₆₄ inches (3.203125" or 81.3mm). Since my Imperial / US Customary measuring tools are mostly going to be marked in sixty-fourths or thirty-seconds or sixteenths anyways, 13/64 is a fairly convenient number and I don't need to figure out what its decimal equivalent is, which means that I can easily do this without any need for a calculator. How's dividing 4,900 by 60?

[Cheesegear]

I will also point out that while 60 is not a particularly convenient divisor in base-2, it's also not a particularly convenient divisor for base-10. 60 is however fairly close to 64, which is a convenient divisor in base-2 - or when using tools marked in base-2 fractions, as is often the case when working with Imperial / US Customary units - and for something like the kobold you're not going to be too far off if you approximate x/60 as x/64.

Who cares? Dividing by 60, is what's correct, whether convenient or not. And it is convenient, because you're using a calculator, because you have access to Excel and/or a calculator, because you're using Photoshop or an equivalent, which means you have access to a calculator, so it shouldn't be hard.

Dividing by 64 means all your models are going to be smaller than they should be, and out of scale with every official product where 1"=5ft, and most home-tutorials that follow the same. The scale is 1/60, not 1/64. I don't know what to tell you.

which means that I can easily do this without any need for a calculator. How's dividing 4,900 by 60?

But you are using a calculator, 'cause you're using a computer to make the cardstock models anyway.