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jimmie · 12-14-07, 08:08 PM

Ok, again thank you for the tutorial.

I have little memory on trigonometry now but trying to beat my brain...
I'm also using a real sliderule to find out the target's speed by law of sine (unfortunatelly typical sliderules don't have "minute" or "knots" scales so it'll take more work other than finding speed by S scale..)

At the end of Example 6 you wrote that you could also calculate diverting tagert speed (via the formula on the tutorial).

I understand convergent course case (is it all angles in the triangle less than 90, right?) but I can't grahpically imagine the triangle and component for the diverting case. (I don't know the English name for the triangle which has an angle larger than 90 deg) I mean... in the converget course case, I imagine a triangle ABC, where A, B, C are angles and a, b, c are sides opposite to the same name angles. so, I imagine:

- uboat is on B, and angle B is scope reading (bearing to target)
- a is uboat speed
- target is on A, angle A is AOB and b is the target speed, found by the sliderule easily (in this case typical sliderule's C/D scale can represent knot because it's just a ratio problem without conversion).

Now, what's it's like for the diverting course triangle? I actually had problem finding out target's speed when I wasn't at collision course..

12-14-07, 08:08 PM	#9
jimmie Lieutenant Join Date: Apr 2005 Location: Japan Posts: 259 Downloads: 0 Uploads: 0	Ok, again thank you for the tutorial. I have little memory on trigonometry now but trying to beat my brain... I'm also using a real sliderule to find out the target's speed by law of sine (unfortunatelly typical sliderules don't have "minute" or "knots" scales so it'll take more work other than finding speed by S scale..) At the end of Example 6 you wrote that you could also calculate diverting tagert speed (via the formula on the tutorial). I understand convergent course case (is it all angles in the triangle less than 90, right?) but I can't grahpically imagine the triangle and component for the diverting case. (I don't know the English name for the triangle which has an angle larger than 90 deg) I mean... in the converget course case, I imagine a triangle ABC, where A, B, C are angles and a, b, c are sides opposite to the same name angles. so, I imagine: - uboat is on B, and angle B is scope reading (bearing to target) - a is uboat speed - target is on A, angle A is AOB and b is the target speed, found by the sliderule easily (in this case typical sliderule's C/D scale can represent knot because it's just a ratio problem without conversion). Now, what's it's like for the diverting course triangle? I actually had problem finding out target's speed when I wasn't at collision course..