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Blowin' Across the Lake: Newsletter Challenge (03/27/07)

Posted March 25, 2007 5:01 PM
Pathfinder Tags: challenge questions

The question as it appears in the 03/27 edition of Specs & Techs from GlobalSpec:

You are in the middle of a quiet lake, 250 feet from the shore, when your boat engine stops working. There is no wind, but you raise the sail of your boat and use an electrical blower to supply force to the sails. The blower power is 30,000 cfm (cubic feet per minute). How long will it take you to reach the shore? You can assume the dynamic coefficient of friction between the boat and the water to be 0.1. The weight of the boat with you inside is 750 lb.

(Update: April 3, 8:35 AM) And the Answer is...

Under these ideal conditions and with the power of the blower alone you will never reach the shore, because the boat will never move. According to Newton's Third Law in order to have the action-reaction effect one of the forces has to be external to the system. In this case, the force of the wind from the blower has to come from outside the boat. Because the blower is on the boat, the force from the blower is opposed by a force in the opposite direction.

This is the reason why you cannot move the boat by pushing the mast of the sails if you are standing inside the boat. You could move it by pushing the mast or the boat itself if you are standing on the shore (be careful, you might fall to the lake when the boat moves). This is also the reason why the air coming from the turbine of a jet is directed toward something that is not part of the plane.

Now, having said the above, we should add the clarification that the conditions described are extreme and very difficult to achieve. I consulted with a friend about this question; he considers that only in very ideal conditions will the boat not move. If, for example, the wind from the blower and the sails are not exactly perpendicular to each other, the net force on the boat will not be zero. The net force on the boat will be zero if, and only if, the sail effectively stops the movement of the air - i.e., the momentum the fan imparts to the air is simply cancelled by the air being stopped by the sails. Of course, it is only the average momentum of the air post-sail that would need to be zero.

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#1

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 1:22 AM

Force on sail results from change in momentum (velocity) of air. The fan causes the air's velocity to change more than the sail does. If they are both acting in opposition the fan will win.

How long will it take you to reach the shore? Longer than if you dropped the sail and just used the fan's propulsion. How long then? We'd need to know boat's area (wetted or frontal?) and change in velocity of air through fan.

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#2
In reply to #1

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 4:44 AM

Hi Davo, I'm thinking the opposite to what I understand you said.

The sail will win because if the fan causes a momentum change of air of Δ, then the sail can ideally cause a momentum change of 2Δ, or in practice a change in momentum of at least greater than Δ in magnitude.

So the boat will travel in the direction that the blower is blowing. At what speed? I'll wait for the sailing types to calculate!

Regards, Jorrie

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#3
In reply to #2

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 10:06 AM

hi jorrie....not being a sailor, i think i agree with davo on this...would not there be a reactive force from the fan that would tend to cancel out the sail?..(maybe the boat goes backwards to the shore, as problem states 250 feet from shore so i assume middle of lake).....better to use the fan like an airboat!

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#5
In reply to #3

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 1:01 PM

Hi dberts, I agree that it is better to use the fan without the sail, like an air-boat, but the challenge requires that we "... use an electrical blower to supply force to the sails".

I assume that the sailor puts up his spinnaker, because that is what makes sense for a tailwind. In this case, the sail causes more velocity change of the air than the fan, hence, the boat will move in the direction of that the fan is blowing in.

The fan causes the average air stream to change form 0 to +v, with a reaction component of -v (backward). The sail causes that same air stream to change from +v to k(-v), where k is some positive constant less than 1. So the fan causes an airspeed change of +v and the spinnaker an airspeed change of:

-kv - v = -v(k + 1)

The reaction component to that is v(k+1). Add the reaction components and you get:

-v + v(k+1) > 0, i.e., a reaction component in the forward direction.

I hope this clears this point at least! Now, how long to reach shore?

Regards, Jorrie

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#6
In reply to #5

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 1:46 PM

hi jorrie....understand your point....and now agree.....thanks......and though the size of the sail is probably not a concern of the problem, i will use your starting point to see if we can reach the shore in time before the beer runs out.

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#12
In reply to #5

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 9:56 PM

Jorrie,

Including one extra word in your quote: "you raise the sail of your boat and use an electrical blower to supply force to the sails". Is it within the spirit of this challenge to say that since I raise the sail, I could then change my mind and drop it again? We'll find out on Monday.

Either way the maximum net change in velocity of air from the fan alone, or the fan and sail combined is ΔVair even if k = 1.

As for actual values, I did a quick Excel spreadsheet with following assumed inputs:

Wetted area = 5m2

Air velocity = 30ms-1

This gave a 'terminal velocity' of ~1.4m/s or 4.5ft/s which was attained after ~4s and it took ~55s to hit the shore.

Jumbucks (#10) raised an interesting point that the force vector arising from ΔVair might be directed sideways so the resultant force arising from ΔVair and the reaction force from the keel would allow the boat to keep accelerating even if it is travelling at the same speed as the air coming out of the blower. With the values I estimated the boat's maximum speed will be well below the air speed, so I don't think this will help.

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#14
In reply to #5

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 3:48 AM

Jorrie,

where is the blower?

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#32
In reply to #5

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 11:32 AM

Jorrie,

You assume that the sail will redirect the wind to the blower, with an assumption that there will be a loss.

My assumption is that the sail will not stop and reverse the airflow.

The sail will only slow the air-stream down as it has to pass by the sides.

As you show it the blower would be able to blow itself up in the air when you mount a parachute on it and fill this up with the blower.

The trust reverser is something else: it really reverses the airflow, and your calculation is correct for that case.

A spinnaker will fill up resulting in a pressure difference between the two sides of the fabric. At the sides you will feel lots of air escaping but it will never go back to where it came from.

A second effect on a sail is the under-pressure created at the backside: this pulls the sail. Your blower will never create this.

Most of the energy that is brought into the airflow is converted into heat.

It is a kind of generator without fuel thing: you assume that the sail will give the power from normal wind and that the blower back trust will be somehow smaller than the resulting force from the sail.

Gwen

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#49
In reply to #32

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 8:18 PM

Hi Gwen,

I agree in normal operation a sail only slows wind down, rather than redirecting it backwards, but this is not normal operation. The blower can direct a 'concentrated' stream of air to any part of the sail that we choose. If it is a spinaker type sail and we direct the air stream to the centre from close range then the air stream will be deflected backwards. This will be incredibly inefficient because the air stream will spread out and have a very low return velocity.

Dave

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#66
In reply to #49

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 3:35 AM

It will be so incredible inefficient that the backthrust of the blower will be greater than the thrust of the sail.

The first propeller planes used the propellor to have more wind flowing over the wings. Later they learned how to design a propeller so that the pull was enough to give the plane the needed speed. From that moment the direct high speed wind from the propellor was not a benefit anymore.

A helicopter is not simply a big blower, it is a revolving system of wings. (with the game of high and low pressure generating the lift)

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#75
In reply to #66

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 5:08 AM

Hi Gwen,

I agree that a helicopter is not simply a big blower. It has systems to maintain stability and enough thrust to lift itself off the ground. But we are talking about a little sailboat that is designed for stability. All we have to do is provide a little, even just a little bit of forward acceleration.

I disagree that Jorrie is saying that a blower could lift itself off the ground using a parachute. This analogy replaces the boat's inertia and hydrodynamic drag with gravity, which is much harder to overcome.

See picture below. If the blower outlet is located inside the control volume of the sail, then the air exiting the blower either accumulates inside the sail profile, ceases to exist or exits backwards (even if it slows down). This is conservation of mass.

The force transfered to the boat by the blower is:

Fb = mass flow rate x ΔVb

where ΔVb is the change in velocity of air through the blower.

The force transfered to the boat by the sail is:

Fs = mass flow rate x ΔVs

The air arriving in the control volume defined by the sail arrives with the same velocity it left the blower, and it exist the control volume backwards so it has had a greater change in velocity, therefore the sail will experiance a greater force than the blower. Obviously the difference is only tiny which I have been saying all along but it still exists. It's not enough to win a boat race, but this only reduces acceleration. Anyway I think I've done this to death and convinced nobody and I think the ANSWER posted next week will be that you use the fan by itself and I have to go now so thanks for the debate and I'll talk to you soon,

Dave

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#295
In reply to #75

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

04/06/2007 3:12 PM

Could you please tell me how I can add a picture into a post

Thanks

Jay

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#358
In reply to #295

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/19/2009 3:54 AM

Hi I found this post useful... http://cr4.globalspec.com/comment/268677 take care... (I was searching for this as well...)

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#97
In reply to #5

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 11:14 AM

Yeah, I don't really see how using the sail would be preferable at all...

your equation -v + v(k+1) = -v + kv + v = kv. being that -1<k<1, meaning v > kv. i would much prefer v to kv... aka, don't use the sail.

Also, this challenge makes no mention of how fast the air is leaving the blower, it seems inconsequential to even state the distance from shore.... so in the spirit of applying variables, im going to say it will take a measure of time i will call T to get to shore, and a boredom factor of B that will be achieved in the process of pushing a boat with a fan.

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#236
In reply to #5

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/03/2007 11:01 AM

Hi Jorrie,

Before I read your reply I might've said: "Drop the sail and point the blower away from the shore". But you prompted the idea of using the sail as a blown wing. Assuming that the boat is perpendicular to the shore line, one should orient the sail at a slight angle off the boat axis. Then one should install the fan ahead of the sail such that it can blow over the "suction side" of the "airfoil".

The resulting force could exceed 1*V. The fan jet entrains a lot of ambient air, which the sail deflects thus adding to the fan momentum. The idea is not original. For example, something similar was proposed by Lookheed for an advance tactical transporter quite a few years ago (obviously, not for sails but for wings). The study could be found on NASA site, for a while at least. And that's not the only one.

Of course, the statement of the problem is missing one parameter, which could be any of the power of the fan, the diameter of the fan or the exit velocity.

But let's assume that the fan surface area is 1 m^2 (about 4 feet diameter, a bit large for a fan). 30,000 cuftmin is approximately 16 m^3/sec. The change in momentum will be 1.2 kg/m^3*16 m^3/sec*16 m/sec = 308 N. At 1 m/s the boat drag will be 0.1*1000 kg/m^3*(1 m/s)^2 = 50 N. So the boat will still accelerate at 1 m/s. The boat weigs 750 lb, which is 337 kg. That gives an initial acceleration of approximately 0.9 m/s^2. Before doing any exercise in calculus, one can say that the boat will reach the shore in less than 250 ft *0.3 m/ft / 1.5 m/s < 60 sec = 1 minute (max speed is less than 2.5 m/s).

By the way, the fan is quite hefty: 16 kg/s/2*(16 m/s)^2 = 4.1 kW. In a minute than means 246000/3600 s= 67 VAh. For a car battery, that's 67/12 = 5.6 Ah, not too much for the average battery.

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#238
In reply to #236

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/03/2007 12:41 PM

I think you may have forgotten the factor of 2 in the power. Also, 500 cu-ft/sec is closer to 14 cu-m/s, which makes quite a difference when cubed. The power delivered to the air is about 1.7-kW (2-kW eletrical power??). However, if we believe the 75-lbs drag (water-weeds?), the blower area would need to be less than about 7.7 sq ft if it was to directly impart the required momentum, which requires more like 4-kW.

You suggest using the sail topology to increase the air's momentum. Turbo-props are one well known way to increase the momentum, but other than cowling round the fan or some kind of venturi (ouch), I don't know of a way that would work with a sail.

Fyz

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#248
In reply to #238

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/03/2007 8:25 PM

Hi Physicist,

You are right, I just don't bother too much with details as exact conversion factors (althought such an attitude proved fatal to one of the Mars Voyagers).

I am rather an engineer. I was happy to see whether that boat can make it to the shore in a reasonable time without sinking because of the weight of the battery.

I was more excited by Jorrie's analysis relative to blowing the sail, which shows clearly that the idea doesn't lead to "pull up your self by the belt". It reminded me about some airplanes that use a blown wing (sail in this case).

I am happy rather to build on ideas than argue them, even by fuelling the craziest ones. In fact, I am happy you took the time to read and answer my "disertation".

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#31
In reply to #2

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 11:05 AM

Anyone who has flown on a commercial jetliner has experienced this effect just as Jorrie explained it -- when the thrust reversers are employed, they slow the airplane by redirecting the exhaust forward.

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#59
In reply to #2

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 2:03 AM

"While boating on a quiet lake" - Does this mean the water was steady and current to be negligible?

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#101
In reply to #1

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 11:55 AM

May I play the devils advocate this time. If you are ON the boat and blow toward the sail which is connected to the boat then you only generate a closed force loop which will NOT move the boat. If you are ON the boat and with or without sail (but better without because of drag) you generate an impulse force due to the air jet generated by the blower and can move the boat with respect to the fixed reference system (the lake surface). The time depends on the speed of air jet not only on its flow. Or may be I am wrong? The drag is propostionnal to the square of speed, to the given form coefficient(0.1) and to the frontal area. Last one can be estimated from the weight since the displaced water volume has the same weight as the boat.

If the impulse force is constant you get a diffrential simple equation wich gives the possibility to estimate the time. If you want I can write it but I thought it is for time being only a matter of qualitative analysis.

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#117
In reply to #101

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 9:41 PM

Hi guest,

Your statement "The time depends on the speed of air jet not only on its flow" is correct. Newton's second law is usually written:

F = ma

But a more useful and general form is:

F = d(mv)/dt

or force is equal to rate of change of momentum. In the case the fan exerts a force on the air to change the air's momentum. The equation can be re-written as:

F = m/dt x dv

F = (mass flow rate) x (change in velocity of air)

As for drag being proportional to frontal area, I think with boats they often use "wetted" area rather than frontal area to calculate drag.

Davo

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#129
In reply to #117

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/29/2007 7:46 AM

I agree with your physics. However, if we accept the physically unlikely premise of the question, once we've overcome friction, the boat will simply accelerate until the change in momentum of the air reduces until it just enough to overcome thefriction. If we arrange to expel all the air from the rear, a stream blown from a 7-sq-ft area (just over 5.4-hp power delivered to the output draught) will get us going to about 0.4 mph. From 6-sq-ft will get us going to 1-mph, and from 4-sq-ft will give 3-mph - so it shouldn't take long at all.

Fyz

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#157
In reply to #1

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/29/2007 1:39 PM

...... the answer my friends, is blowing in the wind. The answer is blowing in the wind.

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#246
In reply to #1

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/03/2007 6:47 PM

If you mount the fan in the bow pointed towards the stern and pull the sail in so the boom is at a slight angle to the long dimension of the boat, the air blowing acros the upwind side of the sail causes a pressure drop which pulls the boat forward. You get the advantage of the propulsion of the fan and the pressure drop of the air movement. That's why a sailbort can sail into the wind. How long to reach shore? --that's up to you mathematicians, i'm only a lit major, but I guess you never sailed.

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#258
In reply to #246

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/04/2007 4:51 AM

Actually, the sum of the forces you get when doing this will (at best) be the same as pointing the blower backwards. This is because the sail will not increase the total momentum of the air - it just redirects it. But there are at least two potential reasons to use the sail anyway - difficulty in fixing the blower to point in the right direction, and controlling the direction of movement when you first start; the latter in particular could be difficult without this, as the question clearly states that there is some unusual drag on the keel (described as a coefficient of friction).

However, I'm worried that, with the blower in the prow and pointing off-side, you might go round in tight circles without ever developing any speed. That would be like starting out upwind from stationary with only the jib; as a sailor, you'd be better placed to comment whether you can do this without the boat turning at least to a reach before you can get going.

Fyz

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#261
In reply to #1

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/04/2007 6:16 AM

I'm not an engineer but as a sometime sailor

1. DONT play the fan on the sail from the stern

2. Sheet the sail in tight i.e. for sailing to windward

3. Play the fan past the sail from the bow on the lee side of the sail . This will generate lift with a forward vector component to move the boat forward

4. The fan will act as a pusher propellor also

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#4

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 10:11 AM

Sounds to me like the proverbial "how high can you lift yourself by pulling on your own boot straps" kinda question.

I'm thinking "equal and opposite" no matter how strong of a wind the fan can generate, will still net a zero movement result.

Ya best put on thick rubber gloves, grap the power chord and "hand over hand" it back to shore.

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#27
In reply to #4

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 9:57 AM

I agree entirely. I believe it breaks one of Newton's Laws.

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#60
In reply to #27

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 2:05 AM

Hi Aqua doc,

I'm not fond of breaking Newton's Laws either, but I believe there is a fundamental flaw in the bootstraps analogy. 'Pulling yourself up by your bootstraps' is a statics problem where the ∑F = 0. Propelling a boat forward using a fan or fan and sail together is a dynamics problem where ∑F = d(mv)/dt. A more useful analogy would be sitting on a skateboard, undoing your bootstraps and throwing your boots in the opposite direction to the way you want to travel.

Dave

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#88
In reply to #60

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 7:34 AM

I agree with Uncle Red.

You have to look at the system as a whole. The sail-boat-blower are one unit. The force of the air pushing against the sail has an equal and opposite component pushing against the blower. Since both the sail and blower are connected to the boat, the resulting force is zero.

You can jump out of the boat with the blower and direct the air stream at the sail. The blower is now no longer connected to the boat. Barring electrocution, you can propel the boat forward. That is, if your coefficient of friction between you and the water is more than the boat and the water.

So, in this scenario, you would use the blower to push the boat until the air stream has no effect. Swim forward and repeat the process.

--Ariel

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#7

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 6:24 PM

I'm sure Jorrie is right as I've seen a similar trick question before but there isn't enough data to produce an unique answer. The electric blower will emit a cone of air and the sail will be curved so the angle at the apex of the cone, the curve of the sail and the distance of the blower from the centre of the sail are all parameters that need to be specified. Also, unless there is a uniform size for all electrical blowers, I should think the velocity of the air emitted should also have an impact. I've no idea what 30,000 cfm feels like but I don't think I could hold above my head pointing due forwards at the spinnaker a blower powerful enough to move a boat so I expect that it would be fixed to the deck and blowing at an upwards angle in order to fill the sail. What is the angle?
Sorry if I sound like a grumpy old (fill in description) but really ...

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#8

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 6:48 PM

There are many many factors missing in this equation to calculate the time/distance formula needed to reach the shore. However this poses a very interesting solution that I believe should be considered.

1.) Being the captain of the vessel and as such having an intimacy with her you would undoubtedly know what she would do under any situation. If you were of a positive mind you would consider where to place the fan...the bow or the stern. If you were of a negative disposition, knowing your ultimate fate...casting yourself to the depths as shark bait would seem the obvious solution with a note scrawled to document your demise.

2.) You are however an excellent seaman with years of study in physics so you ponder the situation making the eventual decision to make ready your mainsail and place the fan to the bow of the vessel directing the flow of air to the stern past the sail. You then adjust the sail to set it slightly so that it catches enough wind from the fan to provide some loft because you know that there will be a difference in pressure created to the sides of the sail. This difference will be enough to begin moving the vessel forward thereby displacing more volume of air as the sail passes through plus the negative pressure that is present to the rear of the fan which is drawing in air from the bow of the vessel will also have some affect on the movement forward. You will set your rudder to reduce friction below the waterline and pray you have enough electric to make the distance to shore.

Once you get to shore and relay your epic journey you will be discredited by the mainstream critics as a quack and eventually lose your craft to the IRS due to some tax evasion scheme you had going from running your company offshore from the vessel you just brought in with a fan and some luck.

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#10
In reply to #8

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 8:17 PM

Having been out on the water a number of times with a failed motor, I would firstly like to congratulate the captain on the foresight to attach an adequate extension cord to the boat to enable the use of a fan. (Of course if the motor was electric, maybe the cord was NOT long enough and has simply become unplugged causing the motor to stop!)

Assuming that the extension cord is not strong enough to use as a line to pull the boat back in to shore, I would place the fan around 30 to 45 degrees off the bow, facing just astern of the mast, so it fills the mainsail and provides the required thrust from both the fan and the sail effects. This should get the boat moving at a reasonable speed so the engine can be plugged in again. (Assumes the fan has it's own length of cord to plug it in!)

Of course the other effect of the mechanically weak, and hence small diameter, extension cord is that it heats up! This is very likely to speed up the melting of the ice block thus dropping the iron mass on the catfish below and making a mockery of a previous challenge where no-one even considered the possibility of a sailboat and its effects!

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#9

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 7:49 PM

I would think a lot would also depend on the design of the fan; straight thru versus a U shape blower such as a leaf blower type where the air does a 180 degree turn or a 90 degree turn(squirrel cage) . Not enought info supplied to determine answer.

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#149
In reply to #9

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/29/2007 10:51 AM

Although I suspect the intended answer is that the boat would go nowhere, that answer is wrong. Getting the boat to go nowhere would require exactly the right size and shape of sail. Anything else, and the boat must go either forward or backward.

If you, as a prudent (but simultaneous nutcase) sailor thought about efficiently powering your boat with a a blower, then you would conclude that the best way to do that would be with a large fan, because these are efficient at moving air at very low pressures. A 30,000 fan requires about 5 HP. You could arrange the fan to blow rearward largely past a close hauled flat sail, or you could arrange it to blow forward into a thrust reverser (a spinnaker). Either way, you could make the boat go pretty quickly wherever you wanted. See post 138 for actual weights.

Granted it's a ludicrous idea, given that a paddle works well in this situation.

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#11

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/26/2007 8:22 PM

30,000 cfm - wow big blower - Blowers usually have a inlet on the side and then blow the air out at right angles to that so you would have to figure that the boat would move at some angle other that straight ahead.

While it would be far more efficient to just point the blower astern and head for shore, the boat would move some what toward the sails and victor toward the inlet of the blower.

The concentrated stream of air would hit the sail and then would spread out with some going up, down, and out both sides (or even out one side if the sails are set at some angle). So the direction you move will be hard to determine without some idea of how the sails are set as well as how large are the sails.

What size outlet on the blower? Where does all this horsepower come from? The weight of the blower makes the 750 lbs sound awful light!

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#147
In reply to #11

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/29/2007 10:41 AM

See my post 138 for real world weights.

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#13

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 3:06 AM

I think you are all assuming the sail is placed on the stern and faces the bow of the boat. In that case I think that Newtons laws will apply (for every action there is an equal and opposite reaction) and the boat is not going to move (assuming the sail catches all the moving air particles.

But if the fan is placed near the bow of the boat and blows on the sail at an angle of about 2 o'clock I think you would get an airfoil affect, similar to a boat sailing into the wind. Plus you would have the force of the fan, well partial force (cos of angle times force) pulling the boat.

I am not smart enough to calculate the speed, but I do believe you would get there.

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#19
In reply to #13

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 5:24 AM

It's just the same if the fan is on the front and blowing across the sail. The boat will not move.

There are those who would disput this, but the commonly accepted theory is that the sail forms an aerofoil and produces a lift vector (on its side of course) at right angles to the chord of the sail. This would try to move the boat in the direction of that lift but the keel encourages the boat to move forwards. This is why you can't sail directly into the wind, at best about 45 degrees to it.

Here's the point though. If the fan is on the bow of the boat, there will be no lift relative to the wind. As soon as the boat starts to move, the movement is cancelled out by the fact that the wind is moving with it.

This isn't entirely relevant but might be worth musing on to make sense of the fan idea. Suppose you are sailing with a 3 knot wind behind you, and a full spinnaker. There's no tide so you're moving forwards nicely. Lets say there's no friction and you're doing 3 knots through the water.

You then start your engine and accelerate to the point where the sail empties and hangs limp because you're outrunning the wind.

What speed are you doing now? 3 knots? But you were already doing 3 knots with a full sail.

If you're motoring at 3 knots and the wind's coming from behind at 3 knots, it won't fill the sail.

Hmmmm.

Andrew

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#15

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 3:57 AM

OK, a cubic foot of air weighs .075 pounds. Moving .075 x 30,000 = 2,250 pounds of force per minute. I'm not and don't play a sailor on TV but it seems that with the blower at 45 degrees to the keel on the foward portion of the boat we could havest some force from the air drawn in 2,250 = 2 A2 = 33.5 pounds force. Surfing Wikipedia indicated that with the wind (blower discharge in this case) at approximately 45 degrees to the sail we could harvest some force from the blower against the sail. No numbers, my guess is half so we are back to 1125 = 2B2 = 23.72 pounds F. That would be aroung 57.22 pounds force along the keel per minute. The side force is retarded by the keel. With a boat that weighs 750 x .1 coef friction we have 75 pounds to motivate. It is too late (or early) for me to be doing the F = ma stuff but it looks like we can get back to the beach before the beer gets warm.

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#16

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 5:00 AM

Hi there. I think that Uncle Red is right. Being a keen sailor, I ought to know this one but I have to confess it's a gut feeling sort of answer.

Assuming the fan was huge and the extension cable long enough - If I put my spinnaker up and turn the fan on, yes, it will fill the sail, but so what?

If the fan is on the boat then there is no reason for the boat to move forwards. It is only if the fan is fixed and the boat can move relative to it that the boat will move.

It's a bit like sitting in a car and pushing the dashboard in the hope that the car will move forwards. If you stand outside the car and push the dashboard, then it might move.

I think a better way might be to see if there is an egg whisk in the galley. That would produce nice propulsion if held over the stern!

Cheers

Andrew

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#17

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 5:06 AM

2 minutes and 49 seconds.

I have a marvelous proof , but......

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#18
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Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 5:10 AM

I correct myself , 2 Min's 32.4 Seconds (ignoring acceleration time)

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#20

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 5:56 AM

Forever. A 30000 cfm blower will destroy the sails.

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#22
In reply to #20

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 6:38 AM

Nice one ! He gets blown over as well , and the new fan boat goes off to the horizon.

Outboard ,sail , and blower - this guy must have life-jacket ,flares,lifeboat....

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#21

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 6:36 AM

I wonder what would happen if the fan was used to suck the air from in front of the sail, thus producing a low pressure area for the sail to be pulled in to. The air from the fan would then be ejected towards the back of the boat, so increasing the pressure behind the sail.............

Should work best with a leaf blower type as described by a previous poster.

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#23

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:35 AM

A blower mounted in the boat would put equal force against the boat and sail. Point the blower at the water to push the boat.

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#24
In reply to #23

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:45 AM

It does not matter where you point the blower the back force is at the blower - pointing it at the water only moves the water and there is not a added back force at the blower.

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#25
In reply to #24

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:48 AM

So it would be a waste of time fitting a propeller on an aeroplane?

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#26
In reply to #25

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 8:27 AM

Just like I said the back force that pushes the blower is the reaction of the blower fins or the porpeller pushing against the air - it does not help to direct the air against water or even a wall for that matter.

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#135
In reply to #26

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/29/2007 9:16 AM

Hovercrafts are an extreme example of the additional reactions that are possible

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#28

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 10:30 AM

It will never get to the intended shore. It might eventually get to the opposite shore if some of the "Wind" misses, of passes through the sail.

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#29
In reply to #28

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 10:35 AM

Or it might fall over the edge.

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#30

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 10:42 AM

Put a sail on the rear of jet engine on a plane. Where will it go? This is like the reverse thrust vanes used to slow planes down upon landing. This is physics. Air coming out of fan equals opposite force on fan body. Since the sail cant catch all of the air, i.e. assuming some efficiency loss, the fan will win and propell the boat backwards, opposite to the sail.

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#33
In reply to #30

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 12:05 PM

But with a sail you can sit it at an angle to deflect the air flow sort of having only one side of the reverse thrust working on your jet engine. Don't forget that on the jet engine the air expands 8 or 10 time during combustion. Also with a blower there is a low pressure side (the inlet).

So with the sail set just so - the boat might just go in a circle? Maybe backwards?

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#34
In reply to #33

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 12:17 PM

"Keep on a'blowin...........you aint going nowhere!" Newton

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#35
In reply to #34

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 1:23 PM

not true...... I've seen Popeye do it many times on the cartoon network.

Given enough spinache and time, he could blow on the sail and travel anywhere in the world.

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#36

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 1:25 PM

The correct answer to the question is; An undeterminable amount of time.

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#45
In reply to #36

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:40 PM

Oh come on, that is the answer to every question if you think of some unknown like "what is the temperature of the water, since this affects the viscous drag". If there are some unknowns then give your answer in terms of the unknowns, or state your assumptions. If every engineer just walked away from a 'challenge' every time there was an unknown we wouldn't even have the wheel!

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#48
In reply to #45

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 8:13 PM

I agree with you Davo . My last post was serious . I found myself calculating 2592.324 /17.01 as a time, having forgotten the hell why . I'd already rattled myself converting (probably without need) to metric . I think others have followed you and done more methodical math. I couldn't make an experiment on this work , but the given blower capacity is a bit extreme.

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#50
In reply to #48

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 8:22 PM

Kris,

I agree about the blower capacity. There is a very useful unit converter at http://online.unitconverterpro.com/

Dave

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#56
In reply to #50

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 10:54 PM

Thanks Dave . If this goes the way of the last 'challenge' I'll be really narked. Kris

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#57
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Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 11:37 PM

With a blower this powerful could we turn the boat upside down, cut a hole in the hull and turn it into a hovercraft? We'd need a pocket knife and sticky tape and things. Part of the airflow would be directed downwards and part directed backwards. We'd reach the shore in 5 seconds flat, after spending 3 weeks trying to making it work. But then we're engineers right?

I thought by now someone would have suggested that the lake might be frozen. Or dry (This happens regularly in Australia).

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#58
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Re: Blowing Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 1:19 AM

Sod it , Id just swim. I watched "A Night to Remember" last night so I now know better than to go in a boat anyway. My next post here will be when they give the 'answer' . If teacher says the Earth's flat , you may as well humour him. 24 hours on the go and I'm still stupid enough to check in here . I could use the spare Min's to do something really useful , maybe balance a pencil on its point. I'm off.

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#37

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 2:05 PM

The boat will go backwards. In this case, if the sail has any effect at all the sail acts like a parachute to slow the boat's motion, deflecting the thrust from the fan sideways (depends on distance and geometry, I would think). The boat would go faster, backwards, with the sail lowered.

If the sail were lowered, the net force supplied by the fan would be about 2400 pounds. But due to friction, the net accelerating force would be 240 lbs.

The boat weighs 750 lbs, which is a mass of (about) 24 slugs . Thus the acceleration would be 10 ft/second-squared. (A = F/M.)

Since the distance to the shore is 250 feet, the net time would be about 7 seconds. (t=sqrroot[2*S/A]).

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#41
In reply to #37

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 5:52 PM

I think I need to revise this. The fan rating is in cfm, not cfs. So the force it supplies would be be 40 not 2400 lbs. This would make the acceleration 0.167 ft/seconds-sqrd. Thus the total time would be about 55 seconds.

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#47
In reply to #41

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:56 PM

How did you calculate this force? Haven't you neglected drag from the water? The boat's "coefficient of friction" was given in the question.

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#38

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 2:16 PM

My guess is infinity. you know the same type of interesting phenomena occurs when you pull on your bootstraps with 750 N of force!

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#39

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 2:32 PM

The blower may move you if you point it away from the shore... like the jets of a plane or rocket. The sails are useful only if the force is external, such as in the case of the wind.

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#40

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 4:07 PM

Looks like a mythbusters challenge.

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#42

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 6:28 PM

A few people were close to correct. the blower has thrust the sail is a deflector. It would only take using the blower and angling the sail to cause the thrust from the blower to be directed in an angle to cause the boat to move to shore. now if the blower was not on the boat but aim at the boat then you could just use it just as you would with nature wind. And then sail back in. it would take some time but you would be on the land then.

Charles

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#193
In reply to #42

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/30/2007 2:50 PM

I'm with you, almost all other posts assume the sail is in fully jibed position, but most dinghies never sail like that.

Assume the sail to be close-hauled, direct the fan from broadside and the air will be deflected rearwards, propelling the boat; the sideways thrust from the fan will be counteracted by the keel/centreboard, and the rudder can compensate for any drift.

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#43

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:05 PM

The overall resultants are zero, excepting irrelevant theories. This boat is going no ware.

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#44

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:19 PM

The force generated by the fan on the sail would be equal to the force exerted by the fan either directly or indirectly on the boat (hull, deck or wherever it interacts with the boat). The only way to generate any lift on the boat's sail then would be to place the fan near the bow of the boat and direct it at an angle greater than 45 degrees to the boat and sail close hauled. I really don't think that any degree of calculation will predict the time it would take to reach shore since there are too many undisclosed variables such as sail area. I'm sure a naval architect could come close though.

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#46
In reply to #44

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 7:47 PM

I do not believe enough information exists to answer this question. Here are a several scenarios (in the extreme):

1) The sail has area zero--Thus the thrust of 30,000 cfm may be calculated, factor in the friction or parasitic drag. You do reach the shore, but not as a result of the sail.

2) You angle the fan at the sail so that air impinges at 90 degrees to the sail cloth. Making an assumption that the sail captures all of the fan generated wind, there can be no movement (it's that lift yourself up by pulling on your bootstrap thing mentioned earlier).

3) You angle the fan straight up and in front of the sail. High velocity air provides a low pressure zone in front of the sail and the buoyancy of the boat is assumed (there is that word again) to be sufficient to oppose the fan generated downward thrust. You do move forward, similar to pointing (a tacking term), but velocity may not be calculated because information is insufficient.

4) Just as in 3) above, you create a low pressure zone in front of the sail, but you also angle the fan such that a vector of its thrust propels you toward the shore. Again, there is insufficient information to calculate the boat movement.

My own experience sailing indicates you might move at up to 3 mph pr 4.4 ft/sec (4 ft/sec for easy math). At that pace, it will take you 62 seconds to reach shore.

If you swim decently, you could get there in about the same or less time (I used to swim the 100yd free in 57 seconds).

Tim

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#51
In reply to #46

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 8:32 PM

I say 6.25 minutes.

hmmmmmm.

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#52
In reply to #46

Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 9:00 PM

I don't agree that if you have 30,000 CFM that you can predict the velocity, it depends on if you are pushing that amount of CFM through a 8 inch pipe or a 36 inch pipe, big difference!

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#54
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Re: Blowing' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 9:16 PM

I agree that you can't predict the velocity, but you can:

- state it as an assumption

- state your answer as a function of the velocity or even as a function of the blower outlet area

- do some research and come up with an estimate

- send in a challenge question of your own that does have all the information

- do nothing

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#53
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Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 9:11 PM

And again you can tack the sail so that it has air passing on both sides with one side getting more air - then the air passing over the side that is pushed out would have a lower pressure (think airplane wing). Still unless you have the FPM versus the EFM (or MPS) you have no idea what the velocity is.

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#55

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/27/2007 10:53 PM

Lets try again: 4.55 minutes

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#61

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 2:06 AM

An electrical blower? I'll just pull on the 250ft extension lead that plugs in somewhere on the shore. Hand over hand at 3 foot a pull at 1 pull every 2 seconds... should not take more than 166.66 seconds

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#62

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 2:37 AM

If you pointed the blower straight down, 30 000 CFM of air thrusting downwards might be enough to off set your personal weight due to gravity and you could flap your wings and fly...

Once you're airbourne, you could flap your arms, yell and attract attention and thus get help from shore.

But be careful about how much attention you do attract, the physics police might arrest you for breaking the laws.

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#63
In reply to #62

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 2:43 AM

Laws of physics generally stick up for themselves pretty well.

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#64

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 2:52 AM

My point exactly my good friend !

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#65
In reply to #64

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 3:01 AM

My good friend,

So how much thrust could it produce?

Your good friend

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Thimk.
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Anonymous Poster
#67

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 3:43 AM

never, of course!

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Anonymous Poster
#68

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 3:50 AM

It would be best to point the fan out the back to move forward but since we have to apply a force to the sail, could you mount the fan at 90 degrees ie facing starboard or port? This would suck air in from the side of the boat but boats dont sail too well sideways I dont think so this takes care of the reaction to the fans force. You use the sail at a 45 degree angle to redirect the airflow out the back of the boat so some percentage of the reaction to this redirection would push boat forward. Still dont know how long as there would be some loss of airflow due to friction on sail or disapation.

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Anonymous Poster
#69

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 3:51 AM

30 000 cfm of air equates to 14 m3/s flow

Because it's not given, I'll say that the blower nozzle is 100mm diameter, thus we have an air velocity of 891 ms-1 [Q=v*a]

Thrust is a force, in simple terms we can use F=m*a

14m3 of air at 20 degC will have a mass of 16.8 kg [1m3 of air = 11.81 Newtons @ 20degC]

Hence we have 15 000 kg of thrust

30000 cfm of air is a lot of air, it takes 15 000 kg of thrust to accelerate it from standstill through a 100mm blower nozzle.

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Power-User

Join Date: Feb 2007
Location: Austraila, the Land Down Under
Posts: 122
#70
In reply to #69

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 4:03 AM

891m/s is almost 3 times the speed of sound. To get the air to exit the blower at this velocity requires a very specialized supersonic nozzle. Try a larger nozzle diameter.

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Thimk.
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Commentator

Join Date: Mar 2007
Location: england
Posts: 88
#71

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 4:21 AM

First, if there is no wind, and presumably no tidal current, how did you get 250 feet from the shore.

Second, how did the crew have the foresight to put a big blower and it's generator or battery, on board.

The answer is the boat would never leave the quayside, the weight of a 30,000cfm blower and generator/battery would instantly sink a boat weighing about 580/600 lb. It would be easier to calculate how quickly it would take to sink 20 feet to the bottom

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gem
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Anonymous Poster
#73
In reply to #71

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 4:30 AM

you got there using the motor before it broke down

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Anonymous Poster
#240
In reply to #71

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/03/2007 2:25 PM

I agree that the boat would sink as a boat that size can not hold a fan that size let alone batteries or generator set.

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Anonymous Poster
#72

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 4:21 AM

Hi All,

Long time lurker first time poster. I believe it is a trick question. I say this because I have observed an experiment based around racing a sausage balloon sled on a wire. When the exhuast from the balloon was reversed and pointed at a large square sail the sled did not go faster, it did not go slower. It went NOWHERE with the thrust of the sail and the balloon canceling each other out.

The protagonist would do better to point his blower aft

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Anonymous Poster
#74
In reply to #72

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 4:46 AM

you are still assuming that the blower is pointing either 0° (ie the bow) or 180°. if the blower was off to one side and blowing air across the boat but then redirected with the sail to 180°, you dont have the equal and opposite problem as the reaction to the airflow is not aligned with the thrust from the fan. in the case of your sausage balloon, if you pointed your outlet at 90° but redirected to 180° your balloon would move forward but also spin around the wire. the boat would have enough resistance hopefully to counter-act this so net velocity in 90° or 270° direction would be zero, but maybe non zero in 0° direction.

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Anonymous Poster
#104
In reply to #74

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 12:33 PM

The issue here is simple if we think of conservation of momentum and the boat and blower as a comlete system.

1. The idea of an air thrust motor blasting air forwards at a flat sail results in zero motion as the net momentum change of the air is zero. The blower takes air and throws it forward which on its own would move the boat backwards (principle of conservation of momentum) but when it hits the sale the air is turned at 90 degrees and exits the boat equally on each side so there is no net momentum change.

2. If we now design a sail which effectively becomes a closed tube around the blower with its open end towards the rear of the boat then all the air exits the rear of the boat. Conservation of momentum means that the boat moves forward.

(I agree it would be much better to have the blower pointing with appropriately designed nozzle and no sail. In fact as the boat is in the middle of the lake if we make the sail very small - approaching zero the boat will happily go backward to the shore)

Now I know that real sails fall somewhere in between these 2 extreme examples so a real world answer is impossible without more information. As others have pointed out you need to know the velocity of the air leaving the rear of the boat. The volume and therefore the mass we are given but there is no way of calculating the velocity of the air from the rear as this is determined not by the blower alone but by the efficiency of the sail in directing the air off the rear of the boat.

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Anonymous Poster
#254
In reply to #104

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

04/04/2007 2:20 AM

I still think the assumption of "air thrust motor blasting air forwards" is wrong, despite having read the "answer", because no-one said the air had to go forwards. you could still direct the air through 90 degrees, but direct it all out the back of the boat

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Anonymous Poster
#76

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 5:20 AM

What size nozzle would you have me use ?

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Anonymous Poster
#77

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 5:32 AM

Just a dumb solution?

Put the fan in the water like a prop. Go where you want.

Must be big battery system to power the fan?

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Anonymous Poster
#78
In reply to #77

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 5:37 AM

unfortunately it says you "supply a force to the sails" so fan must be pointed at sail

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Anonymous Poster
#79

Re: Blowin' Across the Lake: Newsletter Challenge (03/27/07)

03/28/2007 5:45 AM

Newtons law, states that every action has an equal and opposite reaction.

seeing the fan will be mounted to the boat, the force exerted by the fan will be transferred to the mounting on the boat in the opposite direction, so my theory is you are going nowhere.

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