Max Force Wind and Safety

[bewitched]

OK, I shall try to redeem myself....

The example of 5kts at sea level and 5kts at 10,000ft.

Density of air at 20C at sea level is say 1.2kg/m3
Density of air at 20C at 10,000ft is say 0.75kg/m3
Windspeed of 5kts is approximately 2.5m/s

Considering a m3 of air:
At sea level
F = M x A
F = 1.2 x 2.5
F = 3 Newtons

At 10,000ft
F = M x A
F = 0.75 x 2.5
F = 1.875 Newtons

The Force at 10,000ft is two thirds that which is experienced at sea level.

The only difference between these two calculations is the density of the air.

However, I doubt many of us sail at 10,000ft. The variable that drives air density at sea level is air temperature not altitude. Colder air is more dense, has higher pressure, has greater mass and so moves with a greater force for a given windspeed.

OK - now shoot me down

[FrankZ]

Originally Posted by Wotname

OK this is finally getting interesting

Finally? You have pretty high standards.

[Ex-Calif]

Interesting indeed!

Because I get all my best science from Wikipedia I am inserting a link to atmospheric pressure. There is a discussion about pressure lapse rate and temperature lapse rate.

Atmospheric pressure - Wikipedia, the free encyclopedia

Believe it or not air has weight. It is piled on top of us all the way to space. At sea level there is about 14.7 pounds of air piled up on us. At 18,000 feet that weight is about 1/2.

Temperature also decreases with altitude and has it's own "standard" lapse rate.

Atmospheric conditions are almost never standard but for the purposes of discussion we have to pretend it is standard. Otherwise we introduce variables to the problem. Temperature and pressure altitude are the two components that define air density as you correctly point out temperature has something to do with density (but not as much as altitude - or more accurately pressure)

Lot's of smart fellahs went before us and have proven that air is less dense at altitude for standard and non-standard lapse rates. So mass per unit volume of air is less on the lake - It's a given.

The other thing we have to agree upon at some point is that the device measuring the wind is a mechanical device. We could connect the anemometer cups to the gauge with a spring and gears. Most likely it is an electrical generator. Either way it provides a resistive force that we measure. At 5kts indicated the device is providing 5kts of resistive force, equal and opposite to the force blowing on the cups.

The device doesn't care whether you are spinning it with your finger, dipping it it a creek or the wind is blowing it. To register 5 kts the exact same Force has to be apllied

Imagine the cup on the anememeter holds 1 cubic cm of air. Less dense air would have less mass in this same unit of volume.

If we agree that F=mA and we know m is less than A must be higher.

A is a function of velocity and time. Whatever time constant we use has to be consistent so the velocity of the molecules must be higher. if v does not change during the time constant the air is unaccelerated and blowing (or not) steady. v is determined by measuring 2x or more during the time constant and measuing displacement. We practically cannot measure the displacement of air molecules over time with boat instruments.

However we know velocity is higher. It must be because the mass is less, the time constant is the same and the mechanical device is measuring the same Force - 5kts.

Now that we covered chapter 1 - mass & pressure - and chapter 3 - Newton - We have to discuss Bernoulli. Because Bernouli makes the boat go not Newton - unless we drop the boat over a waterfall...

[bewitched]

Originally Posted by Ex-Calif

Interesting indeed!

Because I get all my best science from Wikipedia I am inserting a link to atmospheric pressure. There is a discussion about pressure lapse rate and temperature lapse rate.

Atmospheric pressure - Wikipedia, the free encyclopedia

Believe it or not air has weight. It is piled on top of us all the way to space. At sea level there is about 14.7 pounds of air piled up on us. At 18,000 feet that weight is about 1/2.

Temperature also decreases with altitude and has it's own "standard" lapse rate.

Atmospheric conditions are almost never standard but for the purposes of discussion we have to pretend it is standard. Otherwise we introduce variables to the problem. Temperature and pressure altitude are the two components that define air density as you correctly point out temperature has something to do with density (but not as much as altitude - or more accurately pressure)

Lot's of smart fellahs went before us and have proven that air is less dense at altitude for standard and non-standard lapse rates. So mass per unit volume of air is less on the lake - It's a given.

OK with this so far....

I really must learn how to do this multiple quote thing.

see next post.....

[bewitched]

The other thing we have to agree upon at some point is that the device measuring the wind is a mechanical device. We could connect the anemometer cups to the gauge with a spring and gears. Most likely it is an electrical generator. Either way it provides a resistive force that we measure.

This is where my understanding differs.
I do not believe that the device measures Force, I understand that it measures windspeed, or more correctly Velocity. If it were to measure Force, it would need to be able to establish both the Velocity at which the air is travelling AND the Mass of that air. I cannot see how it measures Mass.

At 5kts indicated the device is providing 5kts of resistive force, equal and opposite to the force blowing on the cups.

No, knots is not a unit of Force, it is a unit of Velocity.

The device doesn't care whether you are spinning it with your finger, dipping it it a creek or the wind is blowing it.

Excactly!! – it doesn’t care how dense the material turning it is, because it doesn’t have any effect on what it is measuring – Velocity. If it was measuring Force, it would need to have some means of determining the density of what was driving it.

Imagine the cup on the anememeter holds 1 cubic cm of air. Less dense air would have less mass in this same unit of volume.

How does it measure this Mass?

If we agree that F=mA and we know m is less than A must be higher.

No, the original question assumed that A was a given remember – regardless of how it is measured - I think we are all familiar with the concept that the greater windspeed the more power in the sails and vice verca, so I think it’s safe to take that variable out of the discussion.
The Force experienced at altitude is less because the Mass of air at altitude is less GIVEN THAT ALL ELSE IS EQUAL (wind speed, air temperate, moisture content, volume of air etc etc). In other words, in the expression F = M x A, we have reduced A and so F must be reduced accordingly.

A is a function of velocity and time. Whatever time constant we use has to be consistent so the velocity of the molecules must be higher. if v does not change during the time constant the air is unaccelerated and blowing (or not) steady. v is determined by measuring 2x or more during the time constant and measuing displacement. We practically cannot measure the displacement of air molecules over time with boat instruments.
However we know velocity is higher. It must be because the mass is less, the time constant is the same and the mechanical device is measuring the same Force - 5kts

Sorry, I don’t follow this at all

[Wotname]

Don't forget that the air slows down (just a tiny bit) when it hits the anemometer cups, that is where the "A" comes into the picture. The velocity before it hits the cup minus the velocity after measured over time gives the acceleration component of F=MA. The mass of the air remains the same (in the particular moment) and the force required to spin the cups can only be derived from the change of velocity (ie. acceleration, + or -).

Not from Wikipedia

[GordMay]

Can anyone explain the difference(s) between the “Heisenberg Uncertainty Principle” and the “Observer Effect”?

[Wotname]

Originally Posted by GordMay

Can anyone explain the difference(s) between the “Heisenberg Uncertainty Principle” and the “Observer Effect”?

Sorry, not me - is this a quiz, is there a prize

[Ex-Calif]

Originally Posted by bewitched

I do not believe that the device measures Force, I understand that it measures windspeed, or more correctly Velocity. If it were to measure Force, it would need to be able to establish both the Velocity at which the air is travelling AND the Mass of that air. I cannot see how it measures Mass.

It has to indirectly measure mass. With no mass the anemometer won't turn. If you struck the cup with a feather a 1m/s versus a brick at 1 m/s the displacement if the anemometer cup would be different.

Originally Posted by bewitched

At 5kts indicated the device is providing 5kts of resistive force, equal and opposite to the force blowing on the cups.

No, knots is not a unit of Force, it is a unit of Velocity.

The anemometer's output could be calibrated in volts or amps.

Originally Posted by bewitched

Excactly!! – it doesn’t care how dense the material turning it is, because it doesn’t have any effect on what it is measuring – Velocity. If it was measuring Force, it would need to have some means of determining the density of what was driving it.

It is insensitive to mass and has no idea what the mass is. Because it is calibrated in knots the total Force at 5 kts has to be the same.

Originally Posted by bewitched

How does it measure this Mass?

It doesn't

Originally Posted by bewitched

No, the original question assumed that A was a given remember – regardless of how it is measured - I think we are all familiar with the concept that the greater windspeed the more power in the sails and vice verca, so I think it’s safe to take that variable out of the discussion.

The original question is whether the boat goes faster at 5 knots indicated at sea level or 10,000 feet. We can't discuss the power from the sails until we understand force. Because the sails react to bernouli's principle.

Originally Posted by bewitched

The Force experienced at altitude is less because the Mass of air at altitude is less GIVEN THAT ALL ELSE IS EQUAL (wind speed, air temperate, moisture content, volume of air etc etc). In other words, in the expression F = M x A, we have reduced A and so F must be reduced accordingly.

This is not entirely correct but if v were the same as at sea level the indicator would read lower.

Because in our example the indicator reads 5 knots at both altitudes F must be the same. If mass is lower than Velocity must be higher.

[sabray]

Originally Posted by Ex-Calif

It has to indirectly measure mass. With no mass the anemometer won't turn. If you struck the cup with a feather a 1m/s versus a brick at 1 m/s the displacement if the anemometer cup would be different.

Okay I couldn't hit mine with a feather. I had to tape the feather to a brick. After a few tries I finally hit the thing. Now it's dangling from the mast head but not turning. I don't think it turned at all it just flopped over. I'm thinking if I had been able to throw the feather that high it wouldn't have flopped so the brick had more force.

[Ex-Calif]

Originally Posted by sabray

Okay I couldn't hit mine with a feather. I had to tape the feather to a brick. After a few tries I finally hit the thing.

Shoulda left it attached to the bird...

[Captain Bill]

I hate to get involved in this but..... Anemometers measure Indicated airspeed just like aircraft Pitot tubes do. A British NACA study testing a cup anemometer in a wind tunnel at know calibrated airspeeds at sea level and 6000 feet showed that an anemometer should be calibrated for altitude to show true airspeed. Those of us who are pilots use a couple of different airspeed, but most commonly indicated and true airspeed. A rule of thumb is that true airspeed increases over indicated airspeed at a rate of 2% per 1000 feet of altitude. This holds pretty well for the lower atmosphere in which this discussion is taking place. The British test showed that an anemometer indicating 4 knots at 6000 feet required 4.6 knots of true wind speed or about 13% low. Not far off of the 12% predicted by the rule of thumb. At 10000 feet it would be off just like a Pitot tube indicator or about 20%. Now the arguement is would a sailboat at 10000 feet and a 5 knot indicated wind speed go slower than at sea level. The answer is no, because a sailboat works on apparent(indicated) wind. The actual (true) wind speed would be closer to 6 knots. This is not the sailing true wind, which is the velocity and direction moving over the ground, this is the indicated speed adjusted for altitude as used in aircraft.

For those who want discuss high altitude takeoffs in aircraft keep in mind F=MA. V=AT, and D=1/2 ATT. The problem is that in large part the force required to accelerate the plane to a speed is based on the true speed of the aircraft, not the indicated speed. At an altitude of 10000 feet an aircraft that took off at 100 Knts at sea level would have to be accelerated to 120 Knots at 10000 feet to get the same indicated airspeed of 100 Knots over the wing. This means that T is 20% bigger given a constatnt accelleration. Since d (distance) increases with the square of time it takes a lot more runway to accelerate to 120 Knts than it does to 100. In addition engine power output drops with increasing altitude. At 10000 feet the available power from the engine on my plane is only about 65% of that at sea level so that reduces the F available for acceleration. It all adds up to a LOT more runway. Temperature and humidity also have an effect and you may hear pilots talk about density altitude. Without getting into that suffice it to say you don't want to try a high weight takeoff at high altitude without a very very long runway.

My 2$ (inflation you know)

[Wotname]

I knew someone would finally mention the significant decrease in power of an aero engine at a high density altitude.

Anyway back to the anemometer; I been thinking that I remember somewhere that it also works by Bernoulli which is why it always spins inn the same direction.

[Ex-Calif]

I have avoided the discussion of the engine performance on purpose because we are really discussing true vs. indicated airspeed.

My point is, and always have been indicated airspeed. Captain Bill is 100% correct. The pitot tube and anemometer work on exactly the same priciples.

He is also correct that at 5kts indicated, TAS is higher at altitude.

He is also correct that at 5kts indicated the sailboat will perform the same regardless of altitude.

At this point a better aircraft analogy is a glider. It removes the confusion tha the discussion of engine performance might cause.

If the glider stalls at 45 kts indicated at sea level, it will stall at 45 kts indicated at altitude.

Imagine the difficulty if the pilot had to memorize a constantly changing graph of stall speeds as altitude changed.

[zydecotoad]

Originally Posted by Ex-Calif

My point is, and always have been indicated airspeed. Captain Bill is 100% correct. The pitot tube and anemometer work on exactly the same priciples.

But the pitot tube and anemometer do not work on the same principles.

The pitot tube measures the dynamic pressure of the air relative to the pitot tube by slowing the air to zero movement and measuring the pressure.

The anemometer balances the drag on the cups facing into and away from the wind. They are built with practically frictionless bearings and remove a negible amount of energy from the air.

Check this link for the equations of motion. See page 33 or so. http://www.ieawind.org/Task_11/Recom...econdPrint.pdf

Todd