How does the angle of the bow of a boat affects its coefficient of drag? PLEASE HELP

If the boat was deep under the water, it is like a wind tunnel test and Froude number, which is related to free surface, is not important. Since your boats are rather bluff, pressure drag is likely to dominate and you will do well with using your Cd formula with cross sectional area and Cd will be (almost) constant at different speeds.

When you are closer to the water and thus making waves, the resistance becomes higher and very much dependent on Froude number. Cd will depend a lot on speed.

Next when part of the hull is above water the displacement starts to reduce and also resistance reduces. Cd will depend a lot on speed.

Thus you should decide which regime you want to measure and then keep the vertical postion of the boat constant.

 

Haha ok, so I calculated wave drag like Ad Hoc suggested and plotted the waved drag against total drag (equal to weight pulling boat forward) for each boat shape. My results look rather bizarre. Again there looks to be a pattern with the three first shapes, but then the last two just go crazy. This is probably due to the fact that they were planning and not totally submerged like the the first three. the first three, however, are also strange because I have negative values, how can I have negative values? I checked my calculations and I am positive that they are correct.

I also made a graph of reynold's number agains total drag. Maybe this is something I could look into. I am thinking of ways of simplifying things. Maybe comparing angle of entrance with reynold's number is a good way to do that. Does a boat which is identical to another but has a different length have the same reynolds number when the same force is pulling it forward? So if a model like mine were to have the same angle of entrance but a different length, and the weight pulling it forward was the same, would both models have the same Re number?

Or I can also take Ad Hoc's advice and compare angle of entrance with speed. Would that be simpler?

 

Ok, that helps. It wasn't deep under water. The first three boats (120, 100, 80) were always submerged but close to the surface, however the other two had the bow out of the water at times. Your explanation about how this affects Cd was very helpful..

 

Perhaps you made an error in typing, or an actual error in calculation.

Did you miss the "1/2" bit?

Since the formula is = 1/2 x rho x WSA x speed^2

If fresh water....roughly guess the temp of the water, if unknown. It would be similar to the air temp. The viscosity can affect your results too.

Could you post the raw data you are using, just for one model and one run, where you get the negative value? So I can double check, ta

That's correct

Remember, you should be plotting your results against Fn. Since this is an indication of something occurring that varies with speed. Until you plot "it", you wont know for sure. The Fn axis is normally the x-axis too, btw.

 

Yeah I forgot to type the 1/2 bit but I did do it. Ok so I got for the 120 degree boat with a 96.6g mass pulling it: average velocity = 1.45m/s, length=0.3577m, visocity = 0.893*10^-6, WSA=0.17177m^2, water density = 1000kg/m^3.

 

Catatau

Ok, I’ve gone through your numbers. I get very similar answers to you, perhaps just rounding up errors, so not worth worrying about for now.
It does indeed indicate a negative value for the wave making, which is clearly not correct.

Thus there is an error in either the WSA calculation or the speed, or some other.

I have looked at model 1, the 120 degree angle, and from you WSA, and the fact you say it is submerged, I work out that the model is roughly 2cm in depth...thus, is this correct, if not, then your WSA area calculation of 0.172m^2 is incorrect.

Lets address that one first.

 

WSA would be the surface area of the whole model right? If it is than my calculation with 3.9cm as depth are correct.

 

Well, your model has the rectangular part of 20x30cm and the triangular part of ½ x 20 x 5.77cm
Which gives a total of 657.7cm^2. Both top and bottom = 1315.4cm^2

The perimeter of one side is 30 + 57.7(for the hypotenuse of the triangle) = 87.7 x 3.9 = 342 x 2 = 684cm^2

So the total is 1315.4 + 684 = 1999.5cm^ = 0.2m^2

So the WSA is incorrect.

 

I agree with the top and bottom part being 1315.4 cm^2 but not the sides being 684cm^2. The side is made of three sides of the rectangle (20x3.9 + 30x3.9 + 30x3.9) which add up to 312cm^2, plus the sides of the triangle which is 11.55 so 11.55x3.9+11.55x3.9 = 90.09

Add up everything 1315.4+312+90.09=1717cm^2=0.1717m^2

 

oh no

Hows this for a bow angle side view !!

 

Hows this for a bow front on view !! 90 feet long its a real strange looking boat thats for sure !!

 

The length of the square section is 30cm. The length of triangular part, the hypotenuse, is 57.7cm.

The triangle, on the centre line is 10cm (half beam) by 5.77cm. So the hypotenuse, is simply square root of 10^2 + 5.77^2 = 11.55cm, ops…I made an error, calculating without my breakfast first!

So total length = 11.6 + 30 = 41.6cm x 3.9xm = 162.2 x 2 = 324.5cm2.

Thus total is 1315.4 + 324.5 = 1639.9cm2 or 0.164m2.

You don’t add the “transom” as such. Since this is perpendicular to the flow and does not contribute to the fwd motion friction, in this sense.

Ok..that’s now done

How did you calculate the speed?

 

Ok so for the speed: I posted a diagram before of how the experiment was set up here it is again, , so the second pulley had a sensor in it that was connected to a computer. The sensor calculated speed and time. I graphed the results (speed x time) and I noticed that it looked somewhat like a terminal velocity graph (increasing velocity until a point where velocity became constant), so I took the points on the graph which were close to a horizontal straight line showing constant velocity and I averaged them out.
Here is an example:

 

Over what distance, from start to finish, was the model dragged?

 

between 2 and 2.5 meters