Curse you, Math for I have not paid attention in class! :(

[Xyphos]

"Why do we have to learn this stuff, we're never going to use it in real life. our lives revolve around the dollar and we only need to know about fractional reserve banking."

--a direct quote from myself, to my former college professor that one semester I took at community college, to which my professor only smiled and continued with the boring lectures.

After choking on my own words, I need help calculating the required height of a nose cone, given a radius of 0.625 and a desired slope angle of 75 degrees, and I'd appreciate the formula used to calculate the answer for future reference/use.

thanks in advance.

[Van Disaster]

It's just a right-angled triangle; you have the angle and the adjacent side ( the radius ), and you want the opposite side ( the height ).

Tan angle = opposite / adjacent -> opposite = adjacent * Tan angle = 0.625 * Tan(75) = ~2.33m

[Xyphos]

+REP

you're awesome! thanks a ton!

[moronwrocket]

Just remember the great Native American princess, Soh cah toa

Sine = opposite/hypotenuse

Cosine = adjacent/hypotenuse

Tangent = opposite/adjacent

[PB666]

Just remember the great Native American princess, Soh cah toa

Sine = opposite/hypotenuse

Cosine = adjacent/hypotenuse

Tangent = opposite/adjacent

Of if you have MS Office Excel

On any right triangle with central angle alpha, distal angle beta, c = length of the hypotenuse, the length of the other two sides a and b are:

[a] =c * SIN(RADIANS(alpha))

=c * COS(RADIANS(alpha))

=c * SIN(RADIANS(beta))

[a] =c * COS(RADIANS(beta))

[a]/ =TAN(RADIANS(alpha))

/[a] =TAN(RADIANS(beta))

[alpha] = DEGREES(ASIN(a/c))

[beta] = DEGREES(ACOS(a/c))

[beta] = DEGREES(ASIN(a/c))

[alpha] = DEGREES(ACOS(a/c))

you can actually cut and paste these (everything from the equals sign right) as long as you Name your the range of your angles and c is the Range name for hypotenuse, and so on.

BTW on blender I find myself doing alot of trig, so. . . . . . . . . . .

An example, Blender produces polygonic cylinders which orient one edge on 0', and you want your collider mesh face to straddle 0' so the parts snap on a flat surface. If the collider mesh is 24-gon this means that each face spans 360'/24 = 15', The cross-section of a radially symmetrical polygon is actually one of the most basic trigonometrical problems, because each face represents the greco-semetic chord. And using the Pythagorean theorem if you know the cord lengths of 180 degrees (diameter) or 60 degrees (radius) you can calculate the length 90, 45, 22.5, 11.25, or the cords for 120, 30, 15, 7.5 degrees without knowing sin or cosine, all you need to know is that a circle is compose of points equidistant from the center and that c² = b² + a²

If Chord 60' = 1 therefore sin 30'=0.5 and the cosine 30 = 0.866025 (definition of isosceles triangle, Pythagorean theorem)
1-0.86602 = exterior radial at 30' =  0.133974; (0.133974[SUP]2[/SUP] + .5[SUP]2[/SUP])1/2 = 0.51763 = chord 30' (Pythagorean theorem)
Since the chord 30'/2 = sin 15' then sin 15' = 0.51269/2 = 0.25882 and since the cos 75 = sin 15 = 0.25882

But I digress, The chord of 15 degrees/2 = sin 7.5 degrees. This solves one problem in that you can rotate your mesh in blender by 7.5 degrees. But in the equation above we see in the second line of code that there is an exterior radial. If your collider mesh is a de-gon version of the texture mesh which with smoothing is virtually circular cross-section with a radius, then the difference between the texture mesh cord r_min and collider mesh cord r_min = r_max (1 - cosine 7.5') = r_max 0.0085.

Have you ever noticed that, for instance, the strut connector starting mesh disappears into the structure you connected to, here is one reason.

This is almost undetectable if the fuel tank is small (r_F1=0.625), but if you rescale your tank to say 7.5 or 10 or 12.5 then r_max can expand to the point that you no longer are able to see maneuvering thrusters (as happens with scale mods of stock parts). This is particularly problematic because the all the symmetry points you have now made flat by shifting 7.5 degrees is also now exactly at the r_min.

Knowing this gives a means for manually changing the radius of the N-gon in the Object mode in blender as long as you know what the maximum scale of modding is going to occur with a part (say 15 meter diameter disk) and the maximum amount of sinking that can be allowed with an attachment.

Lets say our tolerance is 0.05 meters for a maneuvering thruster and r = 0.625M for a cylinder ~ rescaleFactor1, the we can rescaleFactor to 5.88 before the tolerance is exceeded and we have to make a different collider mesh for that part. Another way to put this is that if tolerance is zero then we have to increase size by 0.0086 for each factor over 5.88 so if you want a tank that is F12 radius, then that tanks collider mesh needs to be increased to 1.0045X its original collider mesh radius. Off symmetry maxima part installations would no longer touch the texture unless surface attachment of those parts is designed to sink into the structure. IOW you can neccesarily solve the problem unless with you go for a denser collider mesh, but at least you know the tradeoffs, and you could design mods of strut connectors that are more visible, or thrusters that stand out further.

[Van Disaster]

I have to point out if you're making a cylinder to match stock parts, you might as well make the collider have the same number of sides given there's only 24(? or was it 28, I can't remember now )... there's no cylinder collider so you pretty much either have to use a mesh collider or 12-14 box colliders if you don't need to do anything on the inside.