[HELP] Armour weight vs character weight

[Tema69]

I've had some issues in my current game with the same armour weighing the same for the ST8 75 lb character and the ST16 300 lb character.
Then I stumbled upon the rules in Pyramid's Low Tech II that relate to surface area coverage. They are slightly too complicated in my opinion, considering I just want to adjust armour weight to character weight.

Would character weight divided by 150, multiplied by 2/3 be a reasonable multiplier to armour weight?

A piece of armour that for an average human is 50 lbs would weigh 66 lbs for the 300 lb ST16 character, who has over twice the BL of the 150 lb ST10 average joe.
It doesn't seems too harsh? What do you think?

[Humabout]

I think a rough estimate would to multiply the weight of standard armor by ([Weight]/160)^(2/3). I'm sure someone with more skill in the squishy sciences will correct me, but I'm pretty sure that's the general idea (assuming I don't have fractions inverted or something).

[EDIT]
[Weight] is the weight of the person wearing the armor.

[starslayer]

The issue is that any of this is a function of surface area.

A 5'7" tall man at 250lbs with ST 16 (body-builder) has less surface area then a 6'4" man at 190lbs with ST 12 (basketball player/sprinter/etc).

It would actually cost less in terms of materials (and therefore weight) to armour the 5'9" 250lbs man.

Weight does a lot less for surface area then height does.

[Tema69]

Humabout, I just took the 150 lbs average weight from the pyramid article. What "average weight" we choose is IMO not as important as the formula used. But we seem to agree on that last part :)

Starslayer - the article doesn't take height into account, and I think that risks getting a liittle too detailed for what I need. However, I guess one could make a formula that takes height into account as well?

(character height in cm + weight in pounds) divided by (average height in cm + average weight in pounds) multiplied by (2/3)?

[Ulzgoroth]

Originally Posted by Tema69

Humabout, I just took the 150 lbs average weight from the pyramid article. What "average weight" we choose is IMO not as important as the formula used. But we seem to agree on that last part :)

Starslayer - the article doesn't take height into account, and I think that risks getting a liittle too detailed for what I need. However, I guess one could make a formula that takes height into account as well?

(character height in cm + weight in pounds) divided by (average height in cm + average weight in pounds) multiplied by (2/3)?

The problem is that you seem to be trying to get rid of the 2/3 power in the formula, and doing that is simply not a good idea. Surface area is not proportional to weight.

Multiplying by 2/3 is not remotely okay as a substitute for raising to the 2/3 power.

[Humabout]

Originally Posted by Ulzgoroth

The problem is that you seem to be trying to get rid of the 2/3 power in the formula, and doing that is simply not a good idea. Surface area is not proportional to weight.

Multiplying by 2/3 is not remotely okay as a substitute for raising to the 2/3 power.

This.

If you want to include height as part of it, it's probably somewhere around the average of ([Weight]/150)^(2/3) and ([Height]/[Whatever Average Is])^(3/2). I think it'd be raised to the (3/2). I hate figuring fractional exponents; they are never as intuitive as other aspects of math to me.

[Tema69]

Apologies, english is by no means my first language, and translating my already sub-par math skills from one language to another seems to have led to some misunderstanding.

I appreciate the formulas, Humabout. Using an average of 180cm and 150 lb, and myself as an example (190cm and 180 lb), I'd multiply the weight of my armour by 1.1.

The big 300 lb guy, assuming he's average height, would multiply by 1.3.
Had he been tall (200cm) as well as huge, he'd multiply by 1.4.

It doesn't seem like height has that big of an influence on the formula though.

[Flyndaran]

There are website pages that can figure a person's surface area given height and weight.
Body shapes change as you "stretch" or "compress" them. 7 footers don't have heads that much larger than 4.5 footers after all.

[Humabout]

Originally Posted by Tema69

It doesn't seem like height has that big of an influence on the formula though.

Wait for someone with more biological knowledge than myself to chime in. I'm not entirely convinced height gets a ^(3/2) power. It might be a ^3.

I'd test it against the sites Flyndaran mentioned. See which is more accurate.

[Ulzgoroth]

Originally Posted by Humabout

Wait for someone with more biological knowledge than myself to chime in. I'm not entirely convinced height gets a ^(3/2) power. It might be a ^3.

I'd test it against the sites Flyndaran mentioned. See which is more accurate.

It's not more than a ^2, because it would vary as the square of height if you were scaling the entire thing without the weight constraint.

If I were whipping up a model using weight and height...

thickness = (weight/height)^1/2
area is proportional to thickness*height = (weight*height)^1/2
Note that if height is proportional to the cube root of weight, this comes out to weight^2/3.

So my formula might be Surface Area Multiplier = ((weight/150 lb)*(height/68 inches))^(1/2), on the assumption that the reference wearer is 5'8" in addition to being 150 pounds.

This isn't biological knowledge, just scaling math. Actual humans are more complicated than these calculations. Actual armor is too.