Help With Inboard Shaft Angle

[rich99uk]

Hi please see the pictures as you can see the shaft angle is rather steep well i think so anyway. I wanted to know what angle it should be or how to work it out????
I will be cutting of the keel as this stops it plaining.
Do you think that by makeing the angle less steep it will perform much better?
The boat never originally had a inboard and shaft that was put in by us. But obviously we got the angle wrong so any sugestions or help would be great !!

[rich99uk]

I have decided to completley remove the shaft and keel and putthe shaft in on less of a angle and with no keel.
I am hoping to use a v drive as i cant move the engine eny furthey forward.
i justdont know what the angle should be ??

[CDK]

I won't bother you with math but look at it this way:
With the shaft vertical, the boat goes up like a helicopter. With the shaft at 45 degrees, 50% of the force moves the boat fwd and 50% try to lift it.
With the shaft horizontal, all energy is used for propulsion but you have mechanical problems.
So the trick is to position the shaft as close as possible to the horizontal axis. I think any angle under 20 degrees is acceptable, under 15 is good. But there are also boats with a shaft angle like yours, but they do not plane.

For a 12 degrees angle there are standard parts on the market.

[daiquiri]

And what is the shaft angle in this case? It is not easy to estimate it from the photos (looks like something around 20°, is it so?).

You should give some more info about you boat's characteristics, if you want to obtain a useful advice. Boat waterline length, beam, operative weight, engine characteristics, gear ratio, propeller diameter and pitch is the minimum info you should provide.

Also, have you already done a test run? If yes, what was the max speed obtained at full throttle (measured possibly with gps, in calm water), at what rpm, and what was the boat's trim angle? Did you notice some particular engine fumes behind the boat? Or any excessive vibrations?
That would be useful.

[fcfc]

I am just asking, without any useful answer.

A inclined shaft have generate a vertical component. But up to what point the lift or trim change generated is beneficial to propulsion ?

Given the "profile" of skeg and rudder, is shaft inclinaison that much important ?

Up to what point an angled transmission loss cover more horizontal shaft gain ?

[anthony goodson]

Rule of thumb 5degrees 99.6% 10degrees 98.5% 15degrees 96% of pwer transfered to forward motion .So at 15degrees you have roughly 4% trying to lift your stern and bury your bow

[CDK]

Originally Posted by anthony goodson

Rule of thumb 5degrees 99.6% 10degrees 98.5% 15degrees 96% of pwer transfered to forward motion .So at 15degrees you have roughly 4% trying to lift your stern and bury your bow

I think you need a new thumb.
Vertical and horizontal components follow tan and cot. At 10 degrees the relation is 97/3 %, at 15 degrees 92.8/7.2% and at 20 degrees 87/13%.

[Joakim]

Originally Posted by CDK

I think you need a new thumb.
Vertical and horizontal components follow tan and cot. At 10 degrees the relation is 97/3 %, at 15 degrees 92.8/7.2% and at 20 degrees 87/13%.

If you assume that the thrust is at the direction of the shaft (which is not absolutely accurate), you get the following for 1000 N thrust

deg hor ver
0 1000 0
10 984 174
15 966 259
20 940 342

These are calculated with sin and cos. If you keep the horizontal thrust constant, you can use tan and then get

deg hor ver
0 1000 0
10 1000 176
15 1000 268
20 1000 364

Where did your numbers come from? The perpedincular forces do not sum up to 100%, you need pythagoras.

[daiquiri]

I am affraid that there is no simple formula for the calculation of inclined propeller's characteristics. Cavitation can change significantly the Kt and Kq values for the same shaft angle and the same advance ratio.

There are several models available for props in oblique flows. For non-cavitating regimes Gutsche's model seems to perform pretty well, but it cannot be expressed as a simple multiplication by a single sin, cos, tan or ctg function, since it is based on the blade element theory. Athe general result of that model is that both Kt and Kq increase with shaft angle, since the effective advance ratio J (perpendicular to prop blades) is smaller than in case of shaft aligned with the flow. Therefore, one could expect that the prop efficiency can in some cases even increase with some reasonable shaft inclination. But again, strictly speaking, the model is valid for non-cavitating props only. I remember reading some reports which actually indicate the increase of prop efficiency with shaft angle of attack. If I manage to find them, I'll post the info here. Until then, consider it as just a theoretical result.

Most propellers will work with more or less cavitation developed over its blades, so these results can change considerably. Cavitating propellers experience a decrease in both Kt and Kq as the shaft angle increase, as well as a decrease in efficiency. I am not aware of any validated mathematical model for cavitating propellers in oblique flow, unfortunately. Hopefully someone else will know better.

[CDK]

Originally Posted by Joakim

If you assume that the thrust is at the direction of the shaft (which is not absolutely accurate), you get the following for 1000 N thrust

deg hor ver
0 1000 0
10 984 174
15 966 259
20 940 342

These are calculated with sin and cos. If you keep the horizontal thrust constant, you can use tan and then get

deg hor ver
0 1000 0
10 1000 176
15 1000 268
20 1000 364

Where did your numbers come from? The perpedincular forces do not sum up to 100%, you need pythagoras.

Joakim, good to see I am not the only using math occasionally.
My figures are just approximations, obtained by playing with tan and cot from a table, and of course a calculator. Your results are also not correct.
With a 45 degree shaft angle, both tan and cot are 1, so the force vectors have equal length: 50/50. I think we can accept that without calculations.

At 20 degrees angle, the vectors cannot be 1000/364 or 64 and 36%.

The actual calculation is impossible without data like waterline, displacement and hull shape because the whole package changes position as a result of the vertical vector. But how large that change is depends on these unknown factors.
And then there are also the peculiar models diaquiri has mentioned, where tilting a shaft increases performance.

Like most engineers, I think my figures are reasonably accurate until the opposite is proven.

[Joakim]

Originally Posted by CDK

Your results are also not correct.
With a 45 degree shaft angle, both tan and cot are 1, so the force vectors have equal length: 50/50. I think we can accept that without calculations.

At 20 degrees angle, the vectors cannot be 1000/364 or 64 and 36%.

Please tell me, what would be wrong with my "results"? What do you mean by that comment about 20 deg?

A 1000 N at 45 degrees is equal to 707 N horizontal + 707 N vertical force.

[baeckmo]

Naah, c'mon rich99, give us some numbers as requested by daiquiri; what power at which rpm, what gear ratio aso!! Otherwise these guys will compare their thumbs straight into the eternity.......and no one will remember where it all started!

[bit]

Daiquiri
Therefore, one could expect that the prop efficiency can in some cases even increase with some reasonable shaft inclination. But again, strictly speaking, the model is valid for non-cavitating props only.
I do not understand...
Ciao

[rich99uk]

Ok guys cheers for your comments, Here is some more data for you.
I curentttly running at 80hp at 4500 Rev 2-1 gear box 16 X 11 Prop. full throttle i can do 10 knots and at this point i think i am on the plane as the water rushes through the tunnels under the boat and the back of the boat bottom of the transom is flush with the water. I mean the back of the boat is dry. as soon as it gets a little choppy i come of the plane. i have to be almost open full throttle to get there.
When there is a stong tide it can slow me down to 6 - 7 knots.

But this is all changing, when i change the angle of the shaft i will be runnning with 120HP and using a v drive instead of straight shaft ration of gearbox will be 2:1 i think. And of course there wil be NO keel, Hopefully to make it a planeing hull.
weight of the boat is just over 2 ton so say 2 and a half.
What Other info can i give you Guys ???

[daiquiri]

Originally Posted by bit

I do not understand...

I have outlined the results of Gutsche's model. In my opinion it applies only to props which have not been optimized to account for shaft inclination.
Sorry, but i cannot elaborate more now. It is 00.30 a.m. and tomorrow morning at 5.00 I'll be leaving for a 3-day vacation (boy, I do need it...). So, more to come from me next wednesday. I can't wait to see what will happen in this thread in the meanwhile, but I do have like an idea...