I got the right ans for the wrong reasons, before, or it was confused, not now, tho!
Now I really, really got it clear and straight no doubts, or 2nd guessing about what ans is what.
Clear for anybody, if you are interested in the problem, or doing it also for yourself, and/ or if you also enjoy numbers and puzzles!!!!!!!!
The vector picture with the calculator was right but the ans obtained was really the # of degrees across from 1:30 down to 7:30 O'Clock, this is 4.26 degrees because the vector was calculated in degrees
My firewall, 44mm across, 62.2mm diagonal
So, back to the trig form: Tan ( Angle ) = Opp Side Length/ Adj Side Length
Inserting #s
Tan 4.24 = Opp/62.2
becomes
62.2mm x Tan (4.24deg) = 4.616mm
The numbers are so close and similar, kind of like 2+2=4 but 2x2=4 also.
I would not want to generate anymore confusion tho, my 1st check of 1.4142 x 3deg for 4.24deg on the diagonal only works here because I used equal amounts of Right and Down thrust, if they were different, then a diff trig ratio would be needed.
But, then, again, the graph paper method of adding the vectors shows you the answer in degrees very clearly no matter what pairs of numbers you choose.
Well as soon as I said that I thought, let me try it, so they gave us a picture that showed 6.5 deg Right thrust ( remember that Sea Fury Fun Fighter ?), and I said, what if the down thrust they wanted was even more, tho that was not mentioned.
Like 6.5 Right and 8 deg Down
Adding up the thrust of the vector gives about 10.35 at 140degrees, now this brought up another very interesting situation ( Ha! we dig the hole deeper! ); which is, given the shape of the firewall, you would have to extend the area where your mark is or put a mark out there and draw a line to that point at 140 degrees
This is 10.35 degrees "IN" at 140 degrees and then flip the picture left for down and right if you are doing this like most of us would be looking "AT" the nose of the craft. This would really be looking at the nose 10.35deg "IN" at 220 degrees on the compass face, looking straight at it, not from behind, like you are the captain on a ship looking forward 000 relative.
So this ans would be 10.35degrees IN at 220 degrees down the face of your firewall, doesn't matter where your corner is, it matters what bearing the amount of setback falls on, see?
I was really curious just how this would work out myself.
I used to always just hate it when a teacher in school would pick BAD numbers for a problem that don't leave a good trail to follow, bad numbers like 0,1,2,10, and 45 degrees.
0 makes things disappear! And division by 0 really messes things up!
1 duplicates things!
2, well 2+2 =4 but so is 2x2!!!!!!!!!
10, more duplication with another 0, what happened here!
And if all the examples you ever got turned on 45 degrees, you would miss the challenge of what to do if the problem turned on 55 degrees! Yah it's easy if only on 45 deg just multiply by 1.4142.
So to play with this Sea Fury I picked the unlikely downthrust of 8, to give an answer with a clear range you can see.
You know the guy setting his firewall here would have to draw a plumb line down from the Left and shoot down from the Right at 220 degrees till this intersected the Left plumb line and there set the firewall back the amount to give 10.35 degrees on the diagonal length across that specific firewall.
You know what, I used square numbers for my firewall, and, that made some things very easy.
1: Vector add the thrust correction in degrees: ans in degrees across diagonal
2: Length of diagonal calculated
3: Amount of setback in millimeters at 7:30 O'Clock
Examples of how to calculate the amounts at the other corners around 10 and 4 are given in the 1st examples beginning of post where 44mm x Tan ( 3deg ) = 2.3mm.
*******Something very interesting here, the degree of inclination changes as you turn the hand of the clock around the compass!!!!!!!
Pointing down 180 from the left or right = your down thrust correction angle.
Pointing Left ( from in front of firewall ) = your amt of Right Thrust correction.
Pointing in any other bearing requires calculation for amount of angle, you cannot just simply point at 7:30 O'Clock ( like where your 4th bolt might be on your motor mount ) and assume that's where the max inclination is.
From the Sea Fury example, pointing in 220 degrees true, is the maximum and gives the combination of the two angles 6.5 deg + 8 deg = 10.35 degrees.... on that bearing only, because that is the bearing the resultant vector points in.
I had to think about that to get it all straight, myself.
I kind of did this for me as well as you, I was not clear on how I got the right answer, before or if I got it right at all, now I know it clearly.
Anyone like this puzzle, even see you might make a wood pod and permanently set firewall for thrust correction; this is the right thinking, clear, and the right way to get the answer, finally.
Leonard
