Yep, Dr. Andrew Bauer, who was head of wind tunnel testing at Douglas Aircraft, had no problem with the TM equivalency to DDWFTTW. Here is the TM-model that he demonstrated to fellow aerodynamicists:
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Merged DDWFTTW in a “Real” Wind
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Yep, Dr. Andrew Bauer, who was head of wind tunnel testing at Douglas Aircraft, had no problem with the TM equivalency to DDWFTTW. Here is the TM-model that he demonstrated to fellow aerodynamicists:
last night I took Spork's geometry and offset all the stations by +.75 ft. Fed it back into Javaprop along with a diameter of 17.5 feet, and spinner diameter of 2.6 feet, same airfoils with the Reynolds Numbers adjusted upward to be in line with those reported by javaprop at the design rpm.
I picked the time that the data indicated the highest acceleration during run 1 of .19 m/s2, which would require a net force of ~56 Newtons.
The wind speed at that time was 14.3 MPH, the ground speed of the cart was 26 MPH, the deviation was ~1 degree, so the effective wind speed rounds the same. Giving an airspeed of 11.7 MPH.
So, in humberese;
V_Cart = 26 MPH = 11.62 m/s
V_Head = 11.7 MPH = 5.23 m/s
Cd = 0.28
area = 1.858 m2Density = 1.063 kg/m3
K = 1/2 * Cd * area * Density = 0.27
F_Drag = K * V_Head2 = 0.27 * 5.232 = 7.4 N
P_Drag = F_Drag * V_Cart = 7.4 N * 11.62 m/s = 86 W
Frr = m * crr * g = 295 kg * .013 * 9.81 = 37.6 N
(No humber, V_Cart is not a factor in rolling resistance)
P_rr = Frr * V_Cart = 37.6 N * 11.62 m/s = 437 W
Feeding Javaprop an RPM of 114.5 and an airspeed of 5.23 m/s returns a thrust of 438 N
So,
P_Cart = F_Thrust * V_Cart = 438 N * 11.62 m/s = 5,090 W
P_Cart >> P_Drag + P_rr
What did humber forget?
I'll finish it right later.
I picked the time that the data indicated the highest acceleration during run 1 of .19 m/s2, which would require a net force of ~56 Newtons.
The wind speed at that time was 14.3 MPH, the ground speed of the cart was 26 MPH, the deviation was ~1 degree, so the effective wind speed rounds the same. Giving an airspeed of 11.7 MPH.
So, in humberese;
V_Cart = 26 MPH = 11.62 m/s
V_Head = 11.7 MPH = 5.23 m/s
Cd = 0.28
area = 1.858 m2Density = 1.063 kg/m3
K = 1/2 * Cd * area * Density = 0.27
F_Drag = K * V_Head2 = 0.27 * 5.232 = 7.4 N
P_Drag = F_Drag * V_Cart = 7.4 N * 11.62 m/s = 86 W
Frr = m * crr * g = 295 kg * .013 * 9.81 = 37.6 N
(No humber, V_Cart is not a factor in rolling resistance)
P_rr = Frr * V_Cart = 37.6 N * 11.62 m/s = 437 W
Feeding Javaprop an RPM of 114.5 and an airspeed of 5.23 m/s returns a thrust of 438 N
So,
P_Cart = F_Thrust * V_Cart = 438 N * 11.62 m/s = 5,090 W
P_Cart >> P_Drag + P_rr
What did humber forget?
I'll finish it right later.
We now agree on this since this is what I always stated. What you can do is pick one of your X's as it passes and say that it is to be used as a marker for the speed that the IRF is translating wrt to the earth. It never was my intent nor my contrivance to have the FOR following a painted X around the TM. This was your invention.
We can simply say, "The desired IRF is translating (wrt to the earth) with the same velocity as the top of the belt.", which we do and is what we have always said.
A change in the direction of the belt going around the rollers does not make the IRF change its motion wrt the earth. Maybe you agree to this point but I'm not sure. So the IRF is unaccelerated wrt to any ground based IRF. Objects riding on the top of the belt are motionless with respect to the defined IRF (until they fall off the end.). I think most everyone here "gets" this concept. Llyricist even stated he didn't have your interpretation in mind when he made the statement.
Carts don't ride on frames of reference or projections of them-never have, never will. What everyone here is talking about is that we can make measurements of what the cart is doing wrt the inertial frame of reference defined while its traveling on the belt.
Another little tidbit that you don't seem to get about belts and chains is that they are deformable bodies. Parts of them can be accelerating while other parts aren't, so your assertion is baseless about the whole belt accelerating. The whole belt is moving but only the parts going around the end rollers are accelerating.
Here's your argument succinctly:
Parts of the belt are accelerating so all of the belt is accelerating.
This is generally only true for "rigid bodies" which the belt doesn't happen to be.
It is not at all true for deformable bodies which the belt is.
The only energy to be had (for the cart) is the energy in the differential in the speed of the air mass and the linear belt motion which is exactly what everyone has been saying with the cart on the ground. Objects just sitting on the TM belt and objects just sitting on the earth have no energy wrt to the defined reference frames here. In both reference frames, the air is moving wrt to the defined reference frames for each case in the exact same manner and the "ground/belt top" is motionless wrt to these reference frames.
The TM, or more specifically, the upper surface of the TM belt, is still fully alive and well.
WindGrins
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Subduction Zone
Muse
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And in case ppg does not understand your post Windgrins let me try to simplify it for him.
An IFOR is not a physical object. An IFOR is not attached to a physical object. An IFOR is only a way of observing an object or objects and that frame of reference is moving relative to another IFOR.
The most common IFOR, and it is not a perfect one, has the same velocity and direction as the surface of the Earth. So objects that are at rest relative to the Earth are at rest in that IFOR. Another IFOR used in this discussion has the same direction and speed as the top of the treadmill does. Please note that the top of the treadmill does have a constant speed and direction. The treadmill belt as a whole does not. A simple Galilean transformation of all of the forces on the cart on the treadmill yields the same results as all of the forces on a cart moving at the velocity of the wind where the speed of the wind is the same as the speed of the treadmill. In other words if the cart moves up the treadmill it will go faster than the wind when the wind is at the same speed as the treadmill.
I don't think that will do any good, but I tried. All of the vowels seem to have a very tough time with abstract thought.
An IFOR is not a physical object. An IFOR is not attached to a physical object. An IFOR is only a way of observing an object or objects and that frame of reference is moving relative to another IFOR.
The most common IFOR, and it is not a perfect one, has the same velocity and direction as the surface of the Earth. So objects that are at rest relative to the Earth are at rest in that IFOR. Another IFOR used in this discussion has the same direction and speed as the top of the treadmill does. Please note that the top of the treadmill does have a constant speed and direction. The treadmill belt as a whole does not. A simple Galilean transformation of all of the forces on the cart on the treadmill yields the same results as all of the forces on a cart moving at the velocity of the wind where the speed of the wind is the same as the speed of the treadmill. In other words if the cart moves up the treadmill it will go faster than the wind when the wind is at the same speed as the treadmill.
I don't think that will do any good, but I tried. All of the vowels seem to have a very tough time with abstract thought.
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Christian Klippel
Master Poster
Not to mention the little fact that the earth is also moving around the sun, while the sun is moving around the center of the milky way, which in turn is moving as a whole as well.
It's funny to see that the deniers chose to ignore these little facts of reality, and instead want to make their own pseudo-reality. As if physics is some kind of democracy, where they can vote the laws they like...
Greetings,
Chris
Christian Klippel
Master Poster
How many orders of magnitude is the earth's radius bigger than the treadmill's cylinder radius?
The only thing that's dead are your silly arguments. But keep going on, it's definitely pure comedy gold, only good for entertainment and nothing else.
Humber forgot the torque force necessary to turn the propeller at that airspeed and rpm.
Javaprop shows that torque to be 302 N m. at the prop shaft.
Through the gearing of the drive train that translates to:
302 N m * (23 / 65) = 107 N m
And further, an efficiency penalty of:
107 N m / .91 (efficiency of drivetrain) = 117.6 N m at the axle
That torque opposing the rotation of the wheels will impose an opposing force on the cart of:
F_torque = torque / wheel radius = 117.6 / .343 = 343 N
P_torque = F_torque * V_Cart = 343 N * 11.62 = 3986 W
And Humber's equation:
P_cart = P_Drag + P_rr + P_torque
5089 = 86 + 437 + 3986
5089 = 4509
Uh oh, seems to be a bit of excess power there.
Now dividing all those terms by V_Cart and making a more sensible equation:
Fnet = F_Thrust - F_torque - F_Drag - F_rr = 438 - 343 - 7.4 - 37.6 = 50 Newtons
6 N shy of reality.
But wait! There's more!
If I adjust the pitch, and make it lower by 3.5 degrees, it reduces the torque required to spin it to 263 N m.
263 Nm * (23 / 65) / .91 / .343 = 298 N opposing force from prop torque
Of course, it also reduces the thrust to 401 N, so:
Fnet = 401 - 298 - 7.4 - 37.6 = 58 N
2 N more than reality......
Javaprop shows that torque to be 302 N m. at the prop shaft.
Through the gearing of the drive train that translates to:
302 N m * (23 / 65) = 107 N m
And further, an efficiency penalty of:
107 N m / .91 (efficiency of drivetrain) = 117.6 N m at the axle
That torque opposing the rotation of the wheels will impose an opposing force on the cart of:
F_torque = torque / wheel radius = 117.6 / .343 = 343 N
P_torque = F_torque * V_Cart = 343 N * 11.62 = 3986 W
And Humber's equation:
P_cart = P_Drag + P_rr + P_torque
5089 = 86 + 437 + 3986
5089 = 4509
Uh oh, seems to be a bit of excess power there.
Now dividing all those terms by V_Cart and making a more sensible equation:
Fnet = F_Thrust - F_torque - F_Drag - F_rr = 438 - 343 - 7.4 - 37.6 = 50 Newtons
6 N shy of reality.
But wait! There's more!
If I adjust the pitch, and make it lower by 3.5 degrees, it reduces the torque required to spin it to 263 N m.
263 Nm * (23 / 65) / .91 / .343 = 298 N opposing force from prop torque
Of course, it also reduces the thrust to 401 N, so:
Fnet = 401 - 298 - 7.4 - 37.6 = 58 N
2 N more than reality......
Subduction Zone
Muse
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Want to bet?
Only if it has the radius of the earth, is located along the equator, and has similar mass properties to the earth.
Air power = T*v
438N*5.23ms-1
= 2290W
Shaft Power
114.5rpm = 12rad-s
P_shaft = T*w
=302N.m*12rad
=3624W
Prop Efficiency
Eta = 2290W/3624W =63%
Wheel Power Chain Efficiency= 0.91
= P_ShaftPower /n
= 3624/.91=
= 3989W
No need to bother with the gearing. It does nothing but change the wheel size. No need to bother with the net force.
Air power = 2290W (power available to drive cart)
Wheel Drag = 3989W ( power required to run prop)
Aero Drag = 86W (power to push the cart against the headwind)
Wheel Drag = 437W ( power to push the cart against friction)
Total = 4512W (drag)
Air power to propel cart = 2290W
Available Power = -2222W
If you play around with the efficiency, so lower it, you can get a net force, but then the power deficit increases. That is what you are supposed to see is the gimmick, and not fall for it by adjusting this and that.
And
Wind Power
P = E/t = 1/2*A*p*v^3
Prop diameter = 17.5'
A = 20m^2
v =5.3m/s
P = 1/2*20*1.23*5.3^3
=1831W
The power demand is greater than the entire power of the wind passing through the disk area. Were you hoping to get the rest from Mr Wheel?
ETA:
If you are hoping to raise the "slowing of the air" gambit, you face several problems including Betz's limit on the efficiency.
In this case, the wind speed is 6.39m/s
P = E/t = 1/2*A*p*v^3
P = 1/2*20*1.23*6.39^3
=3209W
The entire power of the wind relative to the ground.
The net force trick is performed by choosing particular air speed, cart speed and efficiency ratios.
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Right, but power in the ground frame is force * v_cart.
Just like it is for aerodynamic drag.
Correct.
Correct. But additionally an effective engine power of 2900 W and max efficiency of 79% Calculated from momentum theory.
Same as I got, within rounding differences.
I know there was no need to use the torque and gearing, you notice that up until then I did use power, I was illustrating that it made no difference to the result.
Though the gearing is necessary to calculate the omega to start with.
Wrong.
Wrong. The power available to drive the cart is force * speed, 5089 W, just like I said.
If you want to talk about air power, then don't multiply the resistive forces by ground speed.
Wrong. And non nonsensical.
Wrong. Does not apply to this situation. That's bluff drag, and it has nothing to do with a propeller driven cart above wind speed.
And the air will lose 2189 J of KE in one second, added to the effective power at the shaft of 2900 J/s , (both from momentum theory) equals the 5089 J/s already calculated from Force * Speed.
That is the power required to keep the cart in motion at Vcart.
Air power = T*v
438N*5.23ms-1
= 2290W
"Javaprop shows that torque to be 302 N m. at the prop shaft."
Shaft Power
114.5rpm = 12rad-s
P_shaft = T*w
=302N.m*12rad
=3624W
Prop Efficiency
Eta = 2290W/3624W =63%
Wheel Power Chain Efficiency= 0.91
= P_ShaftPower /n
= 3624/.91=
= 3989W
Efficiency is 63%
An example from Propdesigner.
v= 5.23m/s
Thrust = 320.3N
Air power = T*v ( Power out = 1675.9W)
320.3N*5.23ms-1
= 1675.2W
Shaft Power (Absorbed power = 2343.9W)
114.5rpm = 12rad-s
P_shaft = T*w
=195.5N.m*12rad
= 2346W
Efficiency ( Efficiency = 71.5%)
eta = Airpower/Shaftpower = 1675.2/2346 = 71.4%
No, the cart's energy is its KE.
The demand to drive the cart
Power = force*velocity = Alldrag*Vcart.
The available (net) power
Power = force*velocity
Air power = T*v
v =velocity in stream tube
Thrust is the force in that tube.
The force to keep the cart as a constant velocity is the sum of the drag components seen at the wheel. The wheel "generator" is a drag force and calculated a V_cart*F_wheel. That is in the submission.
The total power demand is Wheel Power + Drag power
The power to drive the cart to do that, is shy by 2KW.
The entire power of the wind relative to the ground at 6.39m/s wind speed
P = E/t = 1/2*A*p*v^3
P = 1/2*20*1.23*6.39^3
=3209W
The entire power of the wind relative to the ground. To get that you need to bring the wind to 0m/s wrt the ground by 100% efficiency extraction.
Your claim is over-unity.
http://en.wikipedia.org/wiki/Wind_energy
Wind energy is the kinetic energy of the air in motion. Total wind energy flowing through an imaginary area A during the time t is:
E = A . v . t . ρ . ½ v2,
where v is the wind velocity and ρ is the air density. The formula presented is structured in two parts: (A . v . t) is the volume of air passing through A, which is considered perpendicular to the wind velocity; (ρ . ½ v2) is the kinetic energy of the moving air per unit volume.
Total wind power is:
P = E / t = A . ρ . ½ v3
Wind power is thus proportional to the third power of the wind velocity
Wrong as it is, that would be 2189W
(The entire power of the wind brought to a stop at 100% efficient extraction is 3209J/s)
The 2900J is "energy from the wheel" and so over-unity.
Perhaps it will eventually dawn on you that it's all a matter of ratios, which assume the cart is already traveling faster than the wind.
Attachments
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I have no idea what point you imagine you are making.
Congratulations, you can have a prop that is more efficient than the actual prop in that particular circumstance..... Now keep learning.
so what?
Wrong, I already explained why.
And that 2189 J is more than enough to cover all the losses in the cart and add to it's KE to boot.
Wrong, it is the part of the 3989 J of negative work on the cart from the prop torque that isn't lost to heat and aerodynamic losses.
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profpangloss
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The energy from the wheel is really energy being added to the system from the tow vehicle. That energy will be lost as soon as the tow rope is dropped, or the cart tries to accelerate making the rope go slack. I am quite sure this is what Mender discovered when he tried to tow his cart up above WS so he could release the popcorn; and that is of course why we will never see that video!
All the carteer's analysis start off with the cart already going faster than the wind, with energy at the wheel courtesy of being pushed or towed. Then the carteers try to keep that energty and add in the energy from "slowing down the wind" to make the energy budget come out in the black. They never do concede that the energy from the tow vehicle is lost as soon as the cart tries to accelerate.
Subduction Zone
Muse
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That is a lie ppg. There have been analyses given that show the cart will go faster than the wind when it is exactly at wind speed.
And your energy analyses have been shown to be wrong many times over. You should stick to a simpler force analysis. But unfortunately for you force, unlike power, is not frame dependent and it is much harder to hide your errors in a force analysis than in a force analysis.
profpangloss
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And you might want to go back to your day-care centre.
I agree with all of that! It IS possible the top of the belt can be at a constant velocity, but only IF great care is taken to ensure this is so. For medical research there are very special "force treadmills" in use that employ unusually powerful motors and heavy flywheels attached to the rollers in an effort to eliminate the accelerations on the belt due to the non-inertial nature of the belt going around the rollers. It would be interesting to see if the cart can run on one of these specially modified treadmills.
Ordinary treadmill belts do undergo accelerations and those accelerations cannot be isolated from the flat parts of the belt.
Add to this the fact the belt is rotating in the IFOR that the carteers have chosen for their reference frame, and it is undeniable the rotational energy from the belt is frame invariant. The fact is, there are more differences than similarities, between a cart on a treadmill and a cart in the wind. Discounting all of those differences is what is leading you to accept the treadmill equivalency; and that is a big mistake as well as unscientific.
Sure.
The energy of the wind is less than the energy you claim is used by the cart. It is not because of losses, but that the idea is wrong.
An example of barefoot physics promoted two amateurs.
Javaprop applies to propellers driven by an engine. You will not find any sign of the prop shaft taking energy from the wind.
The only source of energy coming in, is the KE of the wind flow, and you are claiming to exceed that. The cart does not work as described, because that would be over unity.
profpangloss
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The gimmick is childishly simple, Humber. Suppose the tow vehicle is putting 10 Watts into the system by towing the vehicle. If the efficiency of the cart from wheels to air is 90%, then the prop will extract 9 Watts from the wind. For a brief instant, the cart has an effective power of 19 Watts and tries to accelerate. Just as soon as it does (or the tow rope is released) the 10 Watts from the tow vehicle is gone. Now the cart has 9 Watts and begins to slow down, constantly losing power as it does, until it is once again well below wind speed. The entire idea is childish and incredibly naive.
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