angle resistance

Discussion in 'All Things Boats & Boating' started by cutting edge, Aug 1, 2017.

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cutting edgeNew Member

Being an amateur I waded through lots of equations trying to find a simple one :
A square panel in water is vertical and at 20degrees to the direction of motion . It adjusts to 40 then 80 degrees. Without calculating values but using x as total drag for 20 , then what factors apply to 40 and 80 ? Is it linear , exponential or hyperbolic ? Simple answers work best with me...

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TANSLSenior Member

Without going into depths, drag is proportional to the projected area of the panel in the direction perpendicular to the velocity vector and that area is proportional to the trigonometric sine of the angle you mention.

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gonzoSenior Member

Drag is also a function of velocity and shape.

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cutting edgeNew Member

So it's about .35 , .65 and 1. Does that describe the compression ratios on the panel such that solid wood at those angles would compress at the contact region? Total compression values would be equal if the wood had the same beam ?

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TANSLSenior Member

The force on the panel, if all other variables are equal, will be effectively proportional to 0.342, 0.643 and 0.985, but I am sorry, I do not understand the questions or the conclusion you would like to reach. Could you explain what it is you want to find out?
I do not know if this can help you, the force exerted by the fluid has a component perpendicular to the panel that is the one that compresses the wood (again multiply by the sine of the angle).

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Mr EfficiencySenior Member

The purpose of the question is not mentioned. Maybe he wants to design a new kind of otter board, but I suspect he is thinking in terms of entry half-angles of ship's bows, and their effect on resistance. Also, no mention is made of speed, which as gonzo mentions, is relevant.

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cutting edgeNew Member

My purpose is to find resistance to a blade or chisel on human skull at different angles ( it's about 40% liquid marrow on the skull ) and wood chisels may be similar. Due to lack of data for wood , then water offers some ideas.
Chisels split perpendicular to motion and compress parallel to motion. A chisel with 45 degree tool angle has tan45degrees vectors of equal force acting at right angle to the splitting wood face. At 20degrees, tan .36 the splitting is about 3 times the compressing effect . At 75 degrees tan 3.73 the splitting is less than one third of compressing , roughly a tenth of the 20degree chisel.

At a given depth of cutting a 20degree blade has about 3 times the contact area compared with 75degree chisel.
Probably the total compression for 20degree is around a third of the 75degree chisel. The bond in the wood at the cutting point may alter this but would be the same for both cases.

Is this a valid idea or are the factors too different / unknown to be meaningful?

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Mr EfficiencySenior Member

Huh ? I hope you don't intend to experiment on actual living people, perhaps you should consult with a mortuary ! In any event, it is all meaningless unless the strike force on your chisel can be controlled precisely. This is bizarre !

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cutting edgeNew Member

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cutting edgeNew Member

Experts say a mulga-wood blade did it. These are bi-convex , 10mm thick and so the strike angle is about 75degrees. A sword can strike at about 20degrees . Resistance at high angle seems to make this cut impossible for wood.

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Mr EfficiencySenior Member

I remember you now !

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Mr EfficiencySenior Member

There is a timber in the outback called Ironwood, for obvious reasons, perhaps obtain some and conduct your own experiments ( not on living people, please).

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Boat Design Net ModeratorModerator

Don't like to direct people to other forums, but in this case, I suspect your questions would be better posted on a forensic pathology forum rather than trying to apply boat or ship hydrodynamics to this problem.

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cutting edgeNew Member

Hullo Mr Mod
I've banged my head silly on all possible forums and impossible academics , pathologists, various Museums and so on. No-one knows nothin . But TANSL talks sense and I value his very useful statements . Thank you Mr TANSL.

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TANSLSenior Member

You are welcome, but I was talking about the force exerted by a fluid on an object submerged in it and what you need is the force exerted by an impact of a solid material over another solid material. I think my explanations are of no use to you at all.
Cheers.

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