The area of a trapeziod (the most common rudder shape): Source: Wikipedia Area = .5 * (a + b) * h Don't forget, Area is in square units; so any multiplication needs to be done "squarely". e.g. 1 square feet = 144 square inches If you are lucky enough to find one to order, you will need these calc's! Find the area you are allowed, then work the equation backwards to find the rudder sizes.
Also to increase the size of a rudder 125% per Big Gun rules, photocopy a picture of the rudder with an enlargement of 113% on a copier.
I tried this and I got some crazy numbers. I have a rudder which is 30/16 at A, 17/16 at B, and 30/16 at H. So I did 30/16*(17/16+30/16) which became 30/16*47/16 which became 1410/16. I divided by 16 to get the inches to it came to 88. Impossible. I referred to an algebra book but the examples never show how to work with fractions only whole numbers. I would appreciate some help here. Thanks
While I am at math... When i need a decimal number for tubes I understand that I take 1/4 and divide 1/4 to get .250 for the size. This is easy as if I know I am working with a 1/4 tube then I know that .25 x 4 =1 so 1/4. But how about 5/32? I get .15625. If I didnt know what that number was to start how would I know to convert it to 5/32. What is the foolproof way to take a decimal number and always get the correct fraction? For example I have been looking at aluminum stuffing tubes that are expressed as .180 id or 4mm. The shafts I buy are 1/8 or 5/32 etc. I had a lot of trouble finding out what the ID decimals equated to as fractions. Any help is great! Thanks
OUCH any one got an aspirin I hate math [] has anyone ever done the math on BC XL Class 3 Rudder (Des Moines)for long class 3 ships that get additional rudder area RUD31XL
Well typically I would simplify first. 1.875 * (1.0625 + 1.875) = 5.507 But, doing it "long hand" would be: = 30/16*(17/16+30/16) = 30/16*(47/16) = 1410 / 256 = 705 / 128 = 352.5 / 64 = 176.25 / 32 = 35 + 1.25 / 32 = 5.507 Now, I noticed you forgot the *.5 I didn't put that in the math but assuming you just forgot.. = 5.507 * .5 = 2.75 sqin
This is a bit of trial and error I'm afraid. There could be an algorithm to it, but I can't figure it out this early in the morning. So... .15625 We know our tubes come in 1/4, 1/8, 1/16, or 1/32 increments. So we multiply the decimal (.15625) by 4, 8, 16, THEN 32. 4 = 0.625 8 = 1.25 16 = 2.5 32 = 5 As soon as you get a whole number, then you have a winner. In this case, the answer is 5/32 The complication is sometimes you get a number that is not divisible by 4/8/16/32 or even 64. In this case, you just take the highest number you can measure. IE if your tape measure can only do 4/8/16/32 then there is no reason to do 64. Taking an example: So we multiply the decimal (.17625) by 4, 8, 16, THEN 32. 4 = 0.705 8 = 1.41 16 = 2.82 32 = 5.64 & the best ruler we have can only read 1/16" increments.... 2.82 / 16 Then the answer is: 2/16; if its OK for the piece to be a little small ... OR you could fudge it while measuring to 2 (AND a little bit) /16 OR 3/16; if its OK for the piece to be a little large.
The "nominial" fraction inch size is not important. What you want is the actual size of a part. Use the decimal.
While statistics is always fun, just do it the easy way. Take a peice of paper that is 1" by X". X being the size of the rudder you are allowed. Cut and tape the paper to be the shape you want. you can't create or lose area this way. Also no math involved.
has anyone ever done the math on BC XL Class 3 Rudder (Des Moines)for long class 3 ships that get additional rudder area RUD31XL [?]
Thats kind of what I am looking at. I was trying to figure out the area of the BC class 6 rudders (and not doing too well with it)
The BC rudders are for class 6 ships with dual rudders if I recall corretly, so why wouldnt they each be 3in^2? Class 6 ships are allowed 4in^2, +50% if they have multiple rudders, 6in^2 total. That split between two rudders would be 3in^2 each. Along the same line whats wrong with 2in^2*150%=3in^2? Why would BC sell rudders that aren't the correct size? That would be pretty dumb from a business prespective.
"While statistics is always fun, just do it the easy way. Take a peice of paper that is 1" by X". X being the size of the rudder you are allowed. Cut and tape the paper to be the shape you want. you can't create or lose area this way. Also no math involved." Bob, I like your way of thinking. Too many numbers............
So based on the math of: 0.5 * (1.875 + 1.0625) * 1.875 = 2.7539 does this mean its 2.75 twice for two rudders which is 5.5 total? If so then its a half inch less than the 50% rule allows for long ships or am I way off base.
Anything wrong with taking a 144 scale rudder graphic, and then using a paint program to strech it 125% on both the X and Y axis?
Depends what you are trying to achieve as a final result, and how much area the scale rudder has to begin with. Do you want a rudder with a specific area? 25% larger than scale? 50% larger than scale? What are you trying to achieve?