math skills needed.

[Helen Keller]

I know the length, so what would the width at the widest point be ????

[l921428x]

800 needed here also.

[LAGC]

OOOH! OOOH! It's been so many semesters since I took trigonometry, let me dust off my Precalculus textbook and give this a try...

Firstly, if we were to split those two right-angles so they form a triangle with the known angle (24°) on the left side, then we know...

180 - 24 = 156 / 2 = 78° each

THEREFORE, if we take 78 away from 90, we are left with 12° angles forming the base of the new imaginary triangle on the right side.

THEREFORE, if we subtract 12 times 2 from 180 we determine that the angle on the far right side opposite of the known angle is 156°.

(Following me so far?)

THEREFORE, we have our new triangle with measurements as follows:

angle A = 156 degrees
angle B = 12 degrees
angle C = 12 degrees

side a = (widest point)
side b = 400m
side c = 400m

Now, using the Law of Cosines to find missing side a (which is the widest point we're after):

a^2 = b^2 + c^2 - 2bc * cosA

a^2 = 400^2 + 400^2 - 2(400*400)*cos(156°)

a^2 = 160000 + 160000 - (320000 * -0.9135454574626)

a^2 = 320000 - (-292334.546388032)

a^2 = 612334.546388032

(square root both sides)

a = 782.518 meters

See? Didn't even have to use the ASS method (where you only know Angle-Side-Side.)

[LAGC]

We better get Krupski in here to check my work, I may be a bit rusty...

[Full Otto]

I know the length, so what would the width at the widest point be ????

You building a garage?

[Altarboy]

We gotta find Helen Keller a girl.

[Altarboy]

And Kudos to LAGC. Good work old boy!

[Helen Keller]

We gotta find Helen Keller a girl.

im surrounded by breeders and idiots.

You building a garage?

newp...... 1/144 scale star destroyer.

[Full Otto]

newp...... 1/144 scale star destroyer.

That sounds pretty cool

[LAGC]

Wait a minute... it looks like I fucked up.

Using the ASS method to figure out the length, I came up with 1753.081 meters.

That means those congruent unknown angles can't be right angles, but rather slightly obtuse. My new calculations came up with 114.326 degrees.

So lets try this again...

180 - 24 = 156 / 2 = 78° each

THEREFORE, if we take 78 away from 114.326 degrees this time, we are left with 36.326° angles forming the base of the new imaginary triangle on the right side.

THEREFORE, if we subtract 36.326 times 2 from 180 we determine that the angle on the far right side opposite of the known angle is actually 107.348 degrees.

(Following me so far?)

THEREFORE, we have our new triangle with measurements as follows:

angle A = 107.348 degrees
angle B = 36.326 degrees
angle C = 36.326 degrees

side a = (widest point)
side b = 400m
side c = 400m

Now, using the Law of Cosines to find missing side a (which is the widest point we're after):

a^2 = b^2 + c^2 - 2bc * cosA

a^2 = 400^2 + 400^2 - 2(400*400)*cos(107.348°)

a^2 = 160000 + 160000 - (320000 * -0.2981746281666314)

a^2 = 320000 - (-95415.881)

a^2 = 415415.881

(square root both sides)

a = 644.527641 meters

That looks better.

(I just checked the solutions on that EMMAths site, and that's the answer they came up with as well: width = 645m, length = 1753m.)

[Helen Keller]

Well then ..

convert that to feet and then to 1/144 scale .

[LAGC]

1753m = 5751.312 feet / 144 = 39.94 feet long
645m = 2116.14 feet / 144 = 14.6954 feet wide

Might be a tight fit for a garage.

[Helen Keller]

how'd i screw that up.

came out to 72.5" long and 33" wide.

[LAGC]

According to Wikipedia, Star Destroyers come in lengths between 1600 and 2915 meters.

https://en.wikipedia.org/wiki/Star_Destroyer

So if you're basing your 1:144 calculations off of that higher figure, it might be closer to what you're estimating.

[LAGC]

Or do you mean inches?

Here's a 1/2700 Imperial Star Destroyer Kit: https://www.revell.com/model-kits/en...t/85-6459.html

It says its dimensions are: 23-1/16" length x 13-7/32" width x 5-25/32" height

If something that small is more what you're looking for.

[Helen Keller]

inches.

i came up with 72.5 length and just over 33 wide ..

i musta fucked up my scale calculations.

[LAGC]

Or to use another example,

Here's a 1/144 Millennium Falcon: https://www.amazon.com/Bandai-Scale-.../dp/B014VTPQ22

One of the reviewers said the model itself is about 9" length and 7" width.

And we all know how much bigger Star Destroyers are compared to Millennium Falcons.

Remember that one episode, where the Falcon was sucked up in the Star Destroyer's massive cargo bay?

Yep, a 1/144 Star Destroyer would be HUGE.

[NAPOTS]

LAGC got the right answer.
I did it differently than he did though.

1753m long x 644.6m wide

@1:144 scale thats 479.3" long x 176.2" wide

probably too big for your garage.

[LAGC]

LAGC got the right answer.
I did it differently than he did though.

Yeah, now that I look at it again, there's really no need to calculate that imaginary triangle on the right or figure out the other angles like I did above.

You can just use the SAS (Side-Angle-Side) method instead, using the known angle (24°), and the known sides (1550m).

Ah well. It was fun anyway.

[Helen Keller]

i found a nifty calculator , I have to use 1/935 scale based upon 5751' long and 2116' wide.

gives me just over 72.6" long and 28.8" wide.

i may possibly have fucked that up as well but that's what im rolling with.