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Suppose you are in a close quarter situation and on a collision course with another vessel. State action? Assuming all other conditions are favorable like calm weather, no current, enough sea room, engines at your disposal, a girl in your bed or whatever you require. Most of you will say, that’s an incomplete question, action will depend on situation to situation and sometimes person to person. That said, there are only five possibilities you can choose from.

**a) Alteration of course with increase of speed**

**b) Alteration of course with decrease of speed**

**c) Alteration of course alone.**

**d) Increase of speed**

**e) Decrease of speed**

And there are no correct or all correct answers. I am not good with guesses but if have to make a guess for a general situation of single target and favorable conditions I will make my preferred order like this (b,a,c,e,d) where “b” being the most preferred action and “d” being least preferred action. WHAT IS YOUR PREFERRED ORDER AND WHY PLEASE SHARE IN COMMENTS. And now you no longer require e- mail id to do that.

let’s compare each of them mathematically.So that we can be sure.

**SPEED INCREMENT VS SPEED REDUCTION**

**Guys one more thing, as question mention close quarter situation so the action which produces a maximum deviation from the initial line of approach will be considered as most effective action.**

Well, the plot says it all. But still, let me explain in brief. Here everything means as usual.

OA- initial relative line of approach

WO- way of own ship

WA- way of another ship

WT2- Increased speed of own vessel from WO

WT1- decreased speed of own vessel from WO(here reverse propulsion)

T1A/T2A- New line of approach after altering speed.

In these examples speed reduction is compared to speed increment of equal magnitude i.e OT1= OT2, just to make it a fair fight. In case 1 __speed__* reduction* produced better results and in case 2

__speed__*produced better results. So mathematically they both are equal and I guess you are right it does depend on situation to situation. But here is my personal opinion.*

__increment__## POINTS TO CONSIDER

- Usually, our ship’s run at full ahead speed and increment of further speed is not possible.

- Even if speed increment option is available, you usually can not increase speed as much as you can decrease speed unless you are moving at very slow speed. So decreasing speed will mostly be more effective

- Increasing speed can take you out of your safe speed zone, which you require to maintain by law at all times. But no rule in COLREGS is against decreasing speed.

- This point I want to tell you in favor of increasing speed, ship move In ahead direction and ship’s momentum is already in the forward direction so decreasing speed might take more time than increasing it.

- If you want to share any other points please comment

I think we have a winner here. Though mathematically both are equal but practically

## DECREASE OF SPEED BEATS INCREASE OF SPEED.

So we have e > d. Do you agree? **One more things guys I have found a unique and easy technique to help you decide which one of above action is better in any situation which I will discuss at end of this post.**

**ALTERATION OF COURSE VS ALTERATION OF SPEED**

Let me quote the rule for you.

“*if there is sufficient sea room alteration of course may be the most effective action to avoid close quarter situation …..”*

Time to test it’s mathematical foundations. If found otherwise I will write a letter to IMO requesting a change. Hope I don’t have to do this. Let’s analyze it.

Here everything means as usual. Some new terms are

WT1/WT2- alteration of speed

WS1/WS2- alteration of course.

Circle WO- all possible alteration of course.

Now I don’t think I have to say anything about it. The plot is shouting alteration of course is a better option loud and clear.

## POINTS TO CONSIDER

- To make speed alteration better then course alteration we have to increase or decrease speed by large value compared to initial speed which is not practical unless you are on a slow speed.

- Speed reduction alters’s initial line of approach from OA to T2A but course alteration changes it by S2A. S2A shows a better deviation. T2A can be better if the vessel can go reverse at more speed than she is presently doing in ahead direction, which is highly unlikely.
- Also, course alterations are much faster than speed alterations.

- FROM SAME, FIGURE IT IS EVIDENT THAT “ALTERATION OF COURSE” IS BETTER THAN “ALTERATION OF COURSE WITH DECREASE OF SPEED”.HOW?

The circle drawn in the figure represents all possible course alteration of own ship at present speed. Now if we reduce speed it will shorten the radius of our circle and hence AS1 and AS2 will approach W as well as OA. So we have a winner here.

**ALTERATION OF COURSE – BEATS – BOTH ALTERATION OF SPEED – AND – ALTERATION OF COURSE WITH DECREASE OF SPEED**.

So the correct trend up to here is (c,b,e,d). **if you tend to differ please comment.**

# FINAL SHOWDOWN

### ALTERATION OF COURSE VS ALTERATION OF COURSE WITH INCREASE OF SPEED

Now the final image is crystal clear here as you see violet lines present’s the maximum deviation from the initial line of approach which represents alteration of course with an increase of speed. Logically speaking, alteration of course with an increase of speed will always be better than just alteration of course because of its speed increase it will prescribe a circle with a larger radius and hence maximum deviation can be obtained. We finally have a winner here.

**ALTERATION OF COURSE WITH INCREASE OF SPEED BEATS EVERYTHING.**

** **So finally we have a preference order to think over. It’s **(a,c,b,e,d).** And it appears most logical to me but I want your reviews and comments on this topic.

# SPECIAL CASE OF A STATIONARY TARGET

I want to talk about Titanic a little more. In my post on collision horizons, we discussed a little about it. Let’s continue, RMS Titanic hit an iceberg on her starboard side. But in spite of OOW’s failure to determine collision horizon, I must appreciate taken by him. As soon as it was reported that an iceberg is right ahead he put the wheel to harder port and turned the engine to full astern. So he chooses an action which is 3^{rd} on our preferred list i.e alteration of course with reduction of speed. And I still think he took the best action to avoid collision. Why? That’s because iceberg was stationary. Let’s analyze the plot

Here the smaller circle represents the alteration of course alone and larger circle represents course altered with an increase of speed. Target is stationary so W and A coincides, so not matter how bigger diameter circle’s we draw, AS1/S2 – these relative lines of approach, provide the same deviation from the initial line of approach.

Increase of speed with alteration of course do provide a better solution here “**mathematically**” but it requires a course alteration by 180 degrees. Also, here increase of speed would rather worsen the situation because it will the length of the relative line of approach, means the target will approach the faster towards own ship, reducing the TCPA.** Here truly the best action is “alteration of course with reduction of speed.”**

# MY LITTLE SECRET

Suppose you are on a collision course with a target and you have only 2 options either increase of speed or decrease of speed. You can calculate which action will be more effective with help of this small technique. Please read carefully because it is most likely that it confuse you. And you need to do all this mentally, plotting is not required here.

THIS TECHNIQUE ONLY WORKS WHEN YOU CAN INCREASE OR DECREASE SPEED BY EQUAL AMOUNT. FOR EX. IT CAN’T TELL YOU THAT INCREASE OF 5 KNOTS WILL BE BETTER THAN DECREASE OF 2 KNOTS OR NOT!! BUT IT CAN TELL YOU IF INCREASE OF 5 KNOTS WILL BE BETTER THAN DECREASE OF 5 KNOTS OR NOT.

STEP 1: Determine current position of the target, and its relative line of approach.

STEP 2: Draw your own course from A in direction of your heading.

STEP 3: Length of the line selected to draw course in the last step must be such that it extend in direction of your course from A proportional to the amount you can increase your speed from the current value and extend it in reverse direction of course from A by the same amount to show same decrease of speed.

STEP 4: From center draw lines to both the ends of your course line drawn in step 3, and check where you have a maximum deviation from the initial line of approach. And that will be your most effective action.

Here in case 1 reducing speed produces more effective action while in case 2 increase of speed produces more effective action

Caution: The above method doesn’t tell you anything else except which is a better action to take. It won’t tell you your new relative line of approach , CPA, TCPA etc.

PROOF

Now look at the proof part of the image, observe O’AW triangle. AO’ and AW represents limiting line of approaches.

*“If O is bisecting O’W then no matter how far or near is C from A on its original line of approach * *O’ will always be closer to C than W”*

So this is why the effect of alteration of speed will be proportional to the angle O’and W will make at C.

THANK YOU FELLA’S ,I HOPE YOU ENJOYED IT.

FUN FACTS: First I decided to add this post in my collision horizon’s post but It would have made it unnecessarily large and boring, which most of it already is. So then I decided to take it as a separate topic. And I am working no how tow to make my articles compact and simple.