Can anybody post the formulas to use instead of using the maneuvering Board…Thanks
Please be a little more specific and someone will be able to help you out.
My guess is with the LOP’s on the Coastal part of the exam, just a guess though.
Skycowboy - Which I could help but I’m a moboard type of guy, Try looking in Bowditch. I’m just the opposite, I have problems with the formulas. I couldn’t figure out how to use the diagrams in Bowditch but didn’t put much time on it either.
If it is the LOP’s remember, Plot the last posit they give you for the ship and the brg to the object. DR the object using your cus and spd then draw the 1st brg from the object from the DR position. This will give you a fix or EP at the last time they gave you in the problem.
I will have to look at my notes. I know anchorman would remember this. You don’t want to use a formula on all of them. I think one of them is called the 7/10th rule. The other one is something similar. You would take the difference (or distance traveled) between the first and second bearing and multiply it by something. Let me look through my stuff. I’ll get back to you unless someone beats me to it.
FYI there are some problems that require using maneuvering boards.
I remember you have to use the SIN and COS functions, but I don’t remember how…
If you are referring to Special Case Bearing problems on the Coasal Nav section, here are a few quick and easy ones:
- 45° / 90° (first bearing is 45° off your heading, second bearing is 90° off your heading, or abeam): Distance run between bearings equals distance off when abeam.
- 22.5° / 45° / 90° (the 7/10 Rule): 7/10 of the distance run between the first and second bearings equals the distance to run between the second bearing (45°) and abeam; and also equals the distance off when abeam. Distance run between the first and second bearings equals the distance off at the second bearing (45°).
- 22.5° / 45° / 90°: Distance run between the first and second bearings equals the distance to run between the second bearing (45°) and abeam; and also equals the distance off when abeam.
- Double Angle (30° / 60°, etc): Distance run between the first and second bearings equals the distance off at the second bearing.
There are other more esoteric ones, but these are easy to remember and apply. The USCG used to always have at least one or two on each test.
I’m a math guy, and when I was prepping for my test I figured out how to do all problems without a mo board (or tables!). Something I learned: if the problem is usually done with a mo board then do it with one. You can see a mistake on paper. With pure trig you better be right.
What exactly are you trying to do?
Course to steer to pass abeam. If your on a course of 349 and light bears 13 degrees off Starboard bow at 10.8 miles What course to steer to pass 2.5 miles abeam to starboard
You add the 13 degrees to the bearing, if it was to port you would subtract .
349+13 = 362 2.5 divided by 10.8 = 2nd Sin = 13.4 362-13.4 = 348.6 or 349 degrees
Easier to do with a mo-board. Or: " some old hippie, caused another hippie to trip on acid". Or: “oh hell another hour of algebra”. What I’m getting at is these are just the same old triangle problems no matter if you use algebra or a mo-board to work them. Personally. I’d use a mo-board, faster and easy to visualize the problem.
Or: Get out your high school math books, ie:
s o/h
C a/h
T o/a
The special case bearing questions never really made any sens to me, period. I’m a geometry/trig guy and even on the board they still really never came out right. I would get comfortable solving questions using the material in Bowditch 2 as it is a reference material you will have in the exam room. The formulas and explanation, although sometimes complicated will be in there. some have tables for these questions. I trained myself to use the formulas in Bowditch 2 and nothing else (unless far simpler) as they’d be available when taking the exam.
The simpler ones, like set drift/distance abeam are far easier on the board. same with intercepting another vessel.