A boat is going 5 kts east, the current is 5 kts south, what’s the boats actual speed/direction?

How to solve a trigonometry word problem with vectors.
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[@barneyadams9869]

It's a trick question. The boat is traveling east at 5kts. The answer is at the beginning of the question. Furthermore any good skipper will have accounted for the current speed ensuring that the boat continued on its bearing travelling east at 5 kts as per the plan. If the boat was travelling SE it would be doing so intentionally. Edit: If not a trick then poorly worded.

[@michaelbaughman8524]

No trick. Yes the boat is making 5 kts east. But the current is also pushing the boat south at 5 kts (which is a pretty stiff current!). So the boat's actual speed and heading made good is the combination of the two components (vectors).

[@barneyadams9869]

@michaelbaughman8524 So the boat isn't going east it's going se?

[@michaelbaughman8524]

@@barneyadams9869 Correct. The bow of the boat is pointing due east, but the direction of travel is SE (135 deg) due to the fact that the water is pushing the boat sideways, to the south.

[@Steve-wz5pz]

@@michaelbaughman8524 Wrong, boy-o. The question states "boat is going 5 kts east". That is the end, for all those who comprehend Plain English.

[@waltergladden15]

The problem should have been stated: a boat is on a heading due east with the motor set at 5 knots.... what is its true course and speed?

[@markj369]

Ir states in the beginning a boat is going east at 5 knots per hour. That is the answer. Everything else is moot

[@oldhardrock2542]

What is (k)not clear in the original problem is the speed and heading of the boat relative to what? As @waltergladden15 points out, the problem states the boat is traveling in a given direction at a given speed. To achieve that heading and speed, what heading and speed through the water should the boat have?

[@Pd-17]

Thank you for your service. I served in the Infantry so navigation was key as well. This question inspired me to pull out some cards I wrote 23 years ago while attending college. I knew it was a vector problem. I just needed the refresher on solving. Great instruction.

[@MrMousley]

Well .. If both speeds are exactly the same, the direction would be South East and the speed would be 5 squared + 5 squared = speed squared (Pythagoras) Speed squared = 25 + 25 Speed squared + 50 Speed = square root of 50 Speed = 7.07 kts

[@CV-uy1jc]

As a Retired Air Force Veteran, Thank You for your Service 👍

[@m.h.6470]

Solution: The directions form a right angle triangle, so Pythagoras helps once again: 5² + 5² = x² x = 5√2 ≅ 7.07 [kts] And the direction is directly in the middle between south and east, as both speeds are the same. So the direction ist directly south-east. If the speeds were different, you could calculate the angle of the triangle with the tan or arctan function.

[@simonharris4873]

The correct answer is directly given in the question. The boat is going 5 knts to the East, that's the answer.

[@johnks6733]

I agree with this answer because the captain would adjust his motor speed & rudder so that his speed would be 5 knts East I Think the question is poorly worded

[@schreckpmc]

Correct. The question is very poorly worded.

[@williammeek4078]

Yes. The better question is what direction is the boat pointed and what is speed through the water.

[@ndailorw5079]

@simonharris4873 I agree. But the instructor gives the correct answer for what he intends but states poorly.

[@ndailorw5079]

@@johnks6733 Agree, the question is poorly stated. But the boat’s heading is 135 degrees, or southeast at 7.07 knots per hour due to the combined and resulting forces of the boat’s speed of 5 knots per hour and its intended heading east, and the current’s speed of 5 knots per hour and a heading south. But the question doesn’t ask at what heading and speed should the boat maintain in order to have a heading east at a speed of 5 knots per hour against a current with a speed of 5 knots per hour and a heading south. That answer would involve the captain’s adjustments to the boat’s speed and heading. Which answer would be an answer not asked for here.

[@ToGoMania19]

I’m enjoying these problems/videos as a flashback to my engineering education. I served as a Navy Nuke on the USS Enterprise- long live Naval Reactors!

[@lanisilvious7098]

TabletClassMath, Thank you for your Service! It's so great that you served

[@danieldennis9831]

Direction/speed is SE ≈7 kts (7.071). When I read the question my first thought was: "I know that this is supposed to be a math question but I am not a sailor or nautical engineer, and so I don't know. And I would not feign to interject my uneducated, landlubber opinion into such a question." THEORETICALLY, if I was told that the current and the wind had equal effect on a boat in these circumstances, it becomes an issue of triangles. On a compass, halfway between S and E is SE and the movement is a right triangle to both forces. The Pythagorean Theorem states that a²+b²+c², the current is a, the wind is b, and the actual speed is c. 25+25=50. 5√2 nautical miles would be the actual traversed distance of the boat in an hour. Or better stated 7.071 KTS or a the ship's crew if asked might say, "We're moving a hair over 7 knots, Captain."

[@bartconnolly6104]

I came up with square root of 50 knots in South East Direction...which is technically more correct at three decimal places. If you were sailing across an ocean and had constant current and wind you could miss an Island you are trying to reach in fog :)

[@mpetersen6]

Call it 7 knots. Provided the current is providing 100 % energy transfer.

[@aeroplod]

@bart connolly6104 Which is why position fixes are taken so adjustments can be made. Practically there is no way of exactly predicting anything when navigation a vessel. This a vector triangle to find one's position rather than find a course to steer so by using ships heading and speed and tidal set and drift one is left with an approximation at best. So, 3 decimal places or even 2 are really not that important in practical dead reckoning. Same with calculating a course to steer ... try asking a helmsman to steer exactly or to expect exactly what the tidal atlas predicts. This is where the art and science of navigation come together. 🤔

[@nyneeveanya8861]

My guess was 135 degrees a 7.1 knots. No pencil so common sense was east is 90 south is 180 going as fast east as south so that’s 45 Which means 90+45=135=directly southeast. Then of course Pythagorean for speed is sq root of 25+25=50. 7x7=49 so remainder less than 1 so speed about 7.1 knots. You explained the steps so well. I just think and can’t say why I do something. I used Pythagorean Theory. Didn’t know why and couldn’t remember what it was called. That’s why you’re the teacher and I could never be. I learn the stuff can apply the stuff but can’t tell you how I know which stuff to use.

[@jeffreyerwin3665]

Here is a simple current problem. Outward Bound Co. does expeditons on what are called "pulling boats." These are 30 foot open boats with 8 oars, two masts, and about a dozen crew members. Often it is necessary to row the vessel as it does not have an engine. Here is the O.B. problem: The Sally M leaves the dock and is rowed up the Wye river for exactly one mile. At that point the instructor looses her favorite hat overboard. However, the crew insists on holding their course up river, and 15 minutes pass before the instructor finally persuades the rowers to reverse course and go after her hat. When the Sally M reaches the hat, the crew notices that they are at the very dock that they started from. The question is: How fast is the current in the Wye river? Assume a zero turn around time, a constant rowing speed, and a constant current in the Wye.

[@jeffreyerwin3665]

Here is a more difficult question for the sailors: Popeye lives in Baton Rouge. He likes to sail his Thistle sailboat the 10 miles down to New Orleans now and then. (He has Olive Oyl come down with a trailer to take him and his sailboat home.) It is a straight shot due south on the Mississippi from Baton Rouge to New Orleans, and the current is a steady 6 Kn south. On some of his trips Popeye notices that the wind speed indicator on land shows 6 Kn out of the north, and on other days that same land indicator shows no wind at all. On which days does Popeye get to New Orleans faster? The days that the land indicator shows wind 6 Kn out of the north, or the days that that indicator shows a flat calm? (Now we will see who is a sailor and who is not!)

[@michaelbaughman8524]

This problem is fairly simple. Old-time sailors had to deal with constantly changing wind and current speed and direction, which they could only estimate. This is called dead reckoning. The only way they could check their actual position was by using a sextant or quadrant and compass and clock to measure the angle above the horizon and direction of different stars and navigation tables to convert to latitude and longitude. They could go for days or weeks without a "fix" if the sky was overcast. Before the invention of reliable sea-going clocks ("chronometers") they could only vaguely guess at the longitude. That is why early maps and charts showed islands and shores at widely different longitudes.

[@ephy1973]

I was thinking it is a 45,45,90 triangle. So we get 5 x square root of 2 kts. And it is 135 degrees.

[@davidjones-vx9ju]

if the boat is going east... it's going east it does not depend on the current

[@robertrichardson1973]

My guess was incorrect. I had the direction correct, but the speed was way off. Thanks for the refreshment of vectors. I had to learn it in college and never used it in the field. I found your channel yesterday and really have enjoyed the videos.

[@m.h.6470]

In this simple case, vectors are completely unnecessary. Just imagine the boat moving 5 length units east due to its speed and in the same time it is moving 5 length units south, due to current. The two lines form a right angle triangle and you can simply calculate the hypotenuse with Pythagoras: x² = 5² + 5² x² = 2 * 5² x = √2 * √5² x = 5√2 x ≅ 7.07 [kts]

[@OddawallWood]

Since the boat travels east, it must move north enough to counter the river´s southward motion.

[@robr3872]

Only to maintain course. If nothing else changes then 135 is SE At roughly 7.5 knots is relative motion and true COG

[@OddawallWood]

@@robr3872 So, does ¨going east¨ mean direction or heading?

[@frederickwelham3829]

@@OddawallWood Direction indicated by compass is the heading, just the way the boat is pointing. Actual direction of travel is course. Using the word "going" in the commentary is a bit confusing.

[@ndailorw5079]

@@frederickwelham3829 …so, is “course” the same thing as drift direction… heading north but on a northeasterly “course,” or “direction,” for instance…?

[@aeroplod]

​@@ndailorw5079Hi, the word course needs qualification hence the confusion. Better to use accepted terms ... The direction of the boat through the water is its heading. The tide direction and speed are its set and drift respectively. The combination of the boat's heading and speed and the tides set and drift will give you ... Course made good (over the ground) or Ground track Speed made good (over the ground. Best not to use 'true course' as mentioned elsewhere as this would mean a true compass heading. If you're interested in all this navigation stuff do a basic RYA theory course at a local evening class. Fun and interesting.

[@Mark_Lacey]

SE @ 7.07kts (set & drift)

[@sydneysmith1521]

I did not study vectors during my formal education. Thank you. Yes, I would like to see more.

[@michaelbaughman8524]

Course made good is SE at 7.07 kts.

[@prfinn3846]

Am I an idiot or does it say that the boat is going east at a speed of 5 kts in the question? So wouldn't that be the answer.....right?

[@jmclane607]

The answer of √50 is only exact at the equator. For the sake of argument, the answer would be very much different, the closer one gets to the pole, the more error one encounters. The answer of √50 at the SW assumes a "flat earth" with no pole.

[@bobcornwell403]

An interesting additional question for this problem would be: If your destination was due East, and your boat could only go 5 kts, how fast would your boat be going due East?

[@aeroplod]

@bobcornwell403 In your question if you steer due north at 5kt you will remain in the same spot over the sea bed. There is no way to proceed towards the east without being taken south of east by the tide. An impossible situation. Anchor up until the tide slows or starts heading towards the east then recalculate and go! A good question though!

[@robr3872]

2.5 kt se

[@robertrichardson1973]

This was my original guess. I didn't even consider the increase of speed.

[@russelllomando8460]

had the 135 but learned the speed, thanks.

[@bobcornwell403]

The course is SE at 5 x sqrt of 2 in knots, or approximately 7.07 kts.

[@larrysullivanl8732]

Have question, shouldn't it b boat wants to go 5kts east, but has current going 5kts south, what is actual speed and direction.

[@williammeek4078]

You might want to rerecord this video. It wasn’t worded well. As worded, the answer is in the question. East at 5 knots. That is known as Speed Over Ground and true bearing (speed and direction as shown on GPS) What you wanted was speed and through the water and heading (direction the ship is pointed) needed to achieve that GPS speed and bearing. Which would be a heading of 135 degrees and speed through the water of 5 times the square root of 2.

[@Geoff_G]

What is the coefficient of friction of the hull?

[@anonracer95]

Hurrah for physics😊

[@RT22-pb2pp]

5 knots east as that is his east speed no matter the current he is fighting current and going 5 knots east. it did not say he is going 5 knots east then encounters 5 knot current his east speed is 5 knots answer is in question. now if he is going 5 knots east and then gets side urrent his east speed may change but it stated 5 knots east

[@UmmCarl]

It's going 5 knots east. Just because there is 5 knot current going south doesn't change the compass headings nor the forward speed of the vessel.

[@johannwatzinger1122]

Much simpler calculate 5 Kts times square root of 2 as the sides are equal.

[@Banshee22068]

I don't have to watch the video to answer this one. The answer is in the first part. The boat is going 5 kts east. It didn't say what direction the boat was pointed or the speed of the thrust. The premise stated the end result.

[@panlomito]

By heart: 5 . SQR(2) knots ( ~ 7.07 ) south east.

[@JoeNarbaiz]

Approximately 7 knots at 135 degrees... southeast direction.

[@carolleenkelmann3829]

Why do "things" pick up speed when affected by a cross current? I would have thought they would lose it, trying to battle against the current.

[@vespa2860]

If the boat was not powered at all it would still have a southward speed of 5kt. Powered east by 5kt - the combination of the two gives the actual speed and direction.

[@bartconnolly6104]

It would lose speed .... depending on where you intend to go. If for example you want to go SE then your course east is pulled south by the current so every 5 east is another 5 south . This can be represented by a triangle 5 west east by 5 North south at the east end of the west east with a hypotenuse of square root of 5 squared plus 5 squared sqrt 50 is about 7.01 If you wanted to go east you would have to aim up current so your triangle would still have the 5 k ots south but you are steering up along the hypotenuse of 5 knots meeting the northern start of this 5 knot representation . This would mean the current squared (25) p,us the speed east squared equals your speed (25) effectively two sides of this triangle are equal meaning you have ZERO speed directly east I.e you can turn straight north into a 5 knot current doing 5 not boat speed and effectively standing still. If a current from the north is equal to your speed You can't move anyway east at all unless you move a bit south as well. If course if you go totally south you won't move east either but your speed will be current p,us boat or 10 knots.

[@michaelbaughman8524]

It would only slow the boat if the boat is headed into the current. A current pushing at 90 degrees to the course would not slow the boat, it would add the southerly component to the speed made good. Think about it this way: if you sail with the current, your boat speed through the water is added to the speed of the water. You are not "battling against the current" if it is pushing you in the direction you want to go.

[@Exciteonium]

I agree this is a badly worded and badly answered question, if the direction was required in degrees it should have been stated in the question. A better up to date question would be :- “ A ship's magnetic compass indicates it is travelling East and its speed is 5 knots, however there is a strong South 5 knot current. What does the ship's Sat Nav indicate its speed (Knots) and direction of travel (in degrees)? BTW no maths required to answer this, just a ruler and graph paper 🙂

[@alidjerdjouri5155]

Hello vectors nice to meet you.

[@mahamunibalakrishnan8263]

7.070 knots going south east.

[@terry_willis]

I don't understand how a current at right angles to the boat would add speed, the vector theory notwithstanding. If the current flow was between 91 and 179 degrees I can under stand how the boat's speed would increase and the fact that it's direction will change. But a right angle current flow would be neutral as far as speed; only the direction would change.

[@EuphTube]

The way this lecture explains it may be misleading to the student because the boat’s intended easterly speed actually stays the same, 5 kn. But, because the cross current exists the boat is also moving southerly at 5 kn, so the two of those together means that over the ground the boat is actually making a 7 kn course in a southeasterly direction. That is the speed increase he’s referring to. Now, If the boat were instead trying to make a straight easterly course over ground, it would lose its easterly speed because it would have to head northeasterly in order to fight the cross current

[@terry_willis]

@@EuphTube Thank you for the explanation. It makes more sense now. (You should be the teacher)

[@sharyno]

Loved hearing about your military background! Very interesting!

[@peterwilliams211]

You are assuming the rudder is straight ahead and not taking into account the windspeed and direction

[@robr3872]

=relative motion

[@robr3872]

Oops forgot to add the boats speed of 5 kt. So 7.5kt

[@chimmed]

5knt east

[@frowe327]

7.07 Kts in a SE direction.

[@Steve-wz5pz]

Well; if the boat is going 5kts E in a 5 kt Southern current - then the boat is going 5 kts E. Period. Solved. Now if the question were "If a boat were going 5 kts/hr east in still water, and suddenly the current is 5 kts/hr south, what’s the boats new speed/direction?" -- well that's much different, right? Still UNSOLVABLE, as the current will push on the side length of the hull, while the engine will be exerting force on the hydrodynamic bow, variables that are unknown. Proof that math people need English people. BADLY. This is why I (and many others) struggled with math in school; imprecise questions that make no effing sense. So there's that.

[@jbeasley2528]

SE at 5 x √2 kts

[@jeffreyerwin3665]

Who is going to give my apparent wind problem a shot?

[@danielmadden9691]

√50 SE

[@garyclark4717]

His position is five knots southeast

[@johnnyllooddte3415]

sq rt 50 se

[@ndailorw5079]

@johnnyllooddte3415 5√ 2 ≈ 7.07 knots per hour southeast, or 135.°

[@jimmckay6396]

5 knots east

[@jeffreyerwin3665]

7.07 Kn SE. Too easy. The 5 kn E obviously means 5 Kn through the water. The "actual speed/direction" also obviously is asking for the boat's speed/direction over the bottom. The speed of the boat nad the current can each be represented by a vector, and vectors can be added tip to tail to produce a new vector that represents the actual speed/course relative to the seabed. In this case, the vectors are at right angles so the x squared + y squared = z squared formula applies. 25 + 25 = 50 and square root of 50 is 7.07

[@davidjones-vx9ju]

we know what he meant... it was not what he said

[@jeffreyerwin3665]

Solve my apparent wind problem. . . . @@davidjones-vx9ju

[@guidosarducci1943]

You would go at 7.07 ne to go 5 e. if You were going to India you would end up in Australia.

[@olivemd]

Got it right after I asked my sailor hubby what kts was. Dumb, I know.

[@danielmadden9691]

√50

[@danielmadden9691]

South East

[@ndailorw5079]

@danielmadden9691 5√ 2 ≈ 7.07 knots per hour southeast, or 135.°

[@raynewport9395]

boat's

[@robertnaremore5347]

Excuse me, but your direction is 135 degrees true.

[@mikerobey5953]

Please word the problem so it matches the math. The speed is 5kts and direction east. You can’t go east and southeast at the same time. If the heading was east and boat speed was 5kts then the math would be true. Work on the question.

[@ndailorw5079]

@mikerobey5953 But would it be incorrect to say that you could go east and south at the same time by going southeast at that time…?

[@rbarker4426]

With your calculation you'll never get to your destination therefore you have to cross cut the current and you actually be traveling less than 5 knots I would take a wild guess and say around 3.5 kph. The more Cross current the longer it takes. If you add in wind and Sea conditions. You're probably only going 2kph haha. Might as well head back home and go a different day. With a tailwind and zero current. Otherwise you can end up at Tin buck ll.lost and out of gas 🤣🤣🤣🤣

[@alastairgreen2077]

Boat's, not boats.

[@robr3872]

This math class not grammar class 🤣🤣🤣

[@ndailorw5079]

@@robr3872 Can’t understand the math if you don’t understand the grammar… the two go hand in hand…

[@rogerfroud300]

Another poorly worded question. Speed relative to what?

[@mikekurtz5868]

Regrettably, his elementary school grammar is decidedly less competent than his math skills. In the context of the math problem the word is spelled "boat's".

[@danielmadden9691]

√50 Kts southeast

[@nickchoporis5901]

Blah, Bkah Blah, Blah Blah Blah ... Get ON WITH IT already .... Answer is 5(2)¹/²kts heading 135°

[@chadrew6]

Southeast at square root of 50 knots