Merged DDWFTTW in a “Real” Wind

[Llyricist]

Which is,of course, irrelevant.This is just a guess, but would that be by a circular argument, starting with the assumption that all the thrust pushing the cart forward can only ever be a proportion of the energy fed into the prop from the wheels?

I may be misjudging it, of course. If the analysis is sufficiently accurate that, for example, it shows the cart is capable of self-starting, then by all means let's see it here. (If it doesn't show that, then he would appear to have missed out something important.)

no it is an "analysis" that simply considers a hypothetical cart, weighing 270 kg with a 5 meter diameter prop, wheels with a crr of under inflated car tires, and the drag of an box with a frontal area of 10 sq meters....

that is going downwind at 9 m/s in a 7 m/s wind. He starts by doing a force and power analysis that proves conclusively that it will accelerate (even with the abysmal specs it has). Though it is flawed in that he assumes maximum theoretical efficiency for the prop, it would still accelerate with more realistic numbers.

Then he attempts the energy bookkeeping in the ground frame, and proceeds to double count, fail to count, mix up energy balance considerations with energy required to do the work, anything that he can to make it look like it comes up short. Kind of reverse Enron bookkeeping.

 

[windgrins]

Symmetric Pylons and lift.

Originally Posted by Dr. Keith:
Sorry to slow things down here, but you guys lost me a bit.

My understanding is that humber thinks the pylons, and possibly the body shape, provide a surface that can convert a variable wind into "tacking" such that the vehicle can travel near the average of that variable wind at multiples of the wind speed.

Actually this is an interesting one.

A symmetric pylon in an airfoil shape can definitely produce lift-obviously not in a DDW path but at an angle of attack. Aerobatic airplanes with symmetric wings do it all the time.

Unfortunately there is only one thing wrong with this idea as the motive power of the cart.

When the cart has a direct tailwind, the pylons are not generating any lift as they are like an airplane wing flying backward. If the wind is off to one side, they generate a marginal bit of drag to push the cart along but clearly would be in "stall" condition with a tailwind.

As the cart approaches WS, they become less and less effective until they produce no lift and and at the most a slight side force in nearly perpendicular to the carts travel direction so this can't get us past WS.

The cart must start exceeding WS at a significant velocity and angle to generate lift in the direction to accelerate the cart.... too late, the cart has to be quite above WS for this to occur. So at best, once the cart arrives quite a bit above WS, a non DDW could generate a minor force on the pylons to help the cart along. Unfortunately for the hypothesis, once there, the cart has substantially already exceeded WS. No joy on this theory!

Regards,

WindGrins

Dr Keith: Windgrins, you smug bastard, stop grinning so much!

Oops!

 

[CORed]

As the record holder, NALSA considers it relevent to share with us any challenge to the record. If they ask us to respond to them regarding any challenge we will.

We will make sure that any challenge to our record is made public.

In the end of course much (or none) of the above will happen because humber couldn't present a coherent challenge if his life depended on it.

JB

In other words, in the unlikely event that humber actually presents a challenge to NALSA, we will be seeing it.

I think it's far more likely that he will send his analysis to some professor or other, quotemine the response, and declare victory.

 

[humber]

I thought the lack of denial didn't augur well.

OED
4.4 Comb. auger-hole, the hole drilled by an auger.

 

[windgrins]

I was surprised by that too!

How anybody can look at that ridiculous excuse for a propeller, and cite that idiot as an authority is beyond me.

I was very surprised that the author of that was willing to post that picture of his prop and then declare since his cart couldn't manage it (even though he had the gear train turning the prop backwards) that DDWFTTW was impossible.

I have a failed cart that I've tried that can't do DDWFTTW also. It's called a "brick" I guess this proves that DDWFTTW impossible!

WindGrins

 

[Llyricist]

Originally Posted by Dr. Keith:
Sorry to slow things down here, but you guys lost me a bit.

My understanding is that humber thinks the pylons, and possibly the body shape, provide a surface that can convert a variable wind into "tacking" such that the vehicle can travel near the average of that variable wind at multiples of the wind speed.

Actually this is an interesting one.

A symmetric pylon in an airfoil shape can definitely produce lift-obviously not in a DDW path but at an angle of attack. Aerobatic airplanes with symmetric wings do it all the time.

Unfortunately there is only one thing wrong with this idea as the motive power of the cart.

When the cart has a direct tailwind, the pylons are not generating any lift as they are like an airplane wing flying backward. If the wind is off to one side, they generate a marginal bit of drag to push the cart along but clearly would be in "stall" condition with a tailwind.

As the cart approaches WS, they become less and less effective until they produce no lift and and at the most a slight side force in nearly perpendicular to the carts travel direction so this can't get us past WS.

The cart must start exceeding WS at a significant velocity and angle to generate lift in the direction to accelerate the cart.... too late, the cart has to be quite above WS for this to occur. So at best, once the cart arrives quite a bit above WS, a non DDW could generate a minor force on the pylons to help the cart along. Unfortunately for the hypothesis, once there, the cart has substantially already exceeded WS. No joy on this theory!

Regards,

WindGrins

Dr Keith: Windgrins, you smug bastard, stop grinning so much!

Oops!

On quibble, the lift and drag of those pylons will probably never have a resultant force vector pointing forward in a relative wind coming from the front quarter.

 

[spork]

I'd be interested to see them, if it's not too much trouble.

Here is the data for the nominal prop design. As is typical, I had to make some minor changes to the planform and thickness at the root and tip to make it manufacturable (e.g. so we could fit the carbon windsurfing mast inside it). The changes should very slightly degrade these numbers - but probably still fall within the limited accuracy anyone would assign to these theoretical results:

vc=18*1.467 !vo (f/s) (this is 18 mph airspeed)
omega=120 !RPM
rho=.00226 !Density (slug/ft^3)1500 ft MSL
vsnd=1116.5 ! Speed of sound (f/s)
rt=8.0 !Tip radius (ft)
rh=0.5 !Hub radius (ft)
nblade=2 !Blade number
ao=5.73 !Lift curve slope of 2-d section
apzl= -5.0 !zero lift angle of attack (deg)
ld=67 !Section max L/D (6412 @350000 Re)
cl=0.9 !CL at max L/D
pflag=1 !1=fixed thrust 2 = fixed power
qdes=50. !design thrust (lb) or power (hp)

Design point:

Thrust= 49.9167099(lbs)
Torque= 124.131477(ft-lb)
Power= 2.83614945 (HP)
etap= 0.845000148

r(ft) chord(ft) incidence(deg) yle(ft) yte(ft)
0.57500 0.18983 80.12490 -0.04746 0.14237
0.72500 0.31718 76.89058 -0.07929 0.23788
0.87500 0.46091 73.75922 -0.11523 0.34568
1.02500 0.61252 70.74371 -0.15313 0.45939
1.17500 0.76452 67.85350 -0.19113 0.57339
1.32500 0.91077 65.09468 -0.22769 0.68308
1.47500 1.04670 62.47044 -0.26168 0.78503
1.62500 1.16922 59.98132 -0.29230 0.87691
1.77500 1.27652 57.62588 -0.31913 0.95739
1.92500 1.36789 55.40093 -0.34197 1.02592
2.07500 1.44340 53.30211 -0.36085 1.08255
2.22500 1.50373 51.32415 -0.37593 1.12779
2.37500 1.54988 49.46123 -0.38747 1.16241
2.52500 1.58312 47.70721 -0.39578 1.18734
2.67500 1.60476 46.05583 -0.40119 1.20357
2.82500 1.61616 44.50084 -0.40404 1.21212
2.97500 1.61860 43.03614 -0.40465 1.21395
3.12500 1.61330 41.65586 -0.40332 1.20997
3.27500 1.60135 40.35435 -0.40034 1.20101
3.42500 1.58373 39.12632 -0.39593 1.18780
3.57500 1.56133 37.96673 -0.39033 1.17100
3.72500 1.53491 36.87091 -0.38373 1.15118
3.87500 1.50513 35.83446 -0.37628 1.12884
4.02500 1.47255 34.85332 -0.36814 1.10441
4.17500 1.43766 33.92371 -0.35942 1.07825
4.32500 1.40088 33.04211 -0.35022 1.05066
4.47500 1.36254 32.20531 -0.34063 1.02190
4.62500 1.32293 31.41029 -0.33073 0.99219
4.77500 1.28228 30.65429 -0.32057 0.96171
4.92500 1.24078 29.93474 -0.31019 0.93058
5.07500 1.19858 29.24929 -0.29964 0.89893
5.22500 1.15579 28.59575 -0.28895 0.86684
5.37500 1.11249 27.97211 -0.27812 0.83437
5.52500 1.06875 27.37649 -0.26719 0.80156
5.67500 1.02457 26.80716 -0.25614 0.76843
5.82500 0.97997 26.26254 -0.24499 0.73497
5.97500 0.93491 25.74113 -0.23373 0.70118
6.12500 0.88934 25.24157 -0.22234 0.66701
6.27500 0.84319 24.76258 -0.21080 0.63239
6.42500 0.79632 24.30298 -0.19908 0.59724
6.57500 0.74857 23.86167 -0.18714 0.56143
6.72500 0.69974 23.43764 -0.17493 0.52480
6.87500 0.64951 23.02992 -0.16238 0.48713
7.02500 0.59748 22.63764 -0.14937 0.44811
7.17500 0.54305 22.25996 -0.13576 0.40729
7.32500 0.48533 21.89613 -0.12133 0.36400
7.47500 0.42290 21.54542 -0.10572 0.31717
7.62500 0.35312 21.20715 -0.08828 0.26484
7.77500 0.27023 20.88070 -0.06756 0.20267
7.92500 0.15414 20.56549 -0.03853 0.11560

Tip Mach Number = 0.0922379121

chord line 0-lift line
Pitch at 7.925' (tip): 18.12 ft 24.90 ft
Pitch at 6.725' 18.32 ft 23.85 ft
Pitch at 5.075' 17.86 ft 22.53 ft


Results for off-design performance for ct, cp, cq, and eff as function of advance ratio (jj):

jj 1.2378 0.9902 0.8252 0.7073 0.6189
ct 0.0178 0.0516 0.0731 0.0874 0.0846
cp 0.0254 0.0577 0.0713 0.0775 0.1102
cq 0.0040 0.0092 0.0114 0.0123 0.0175
eff 0.8700 0.8800 0.8400 0.7970 0.4800

jj = v/nD
ct = thrust/(ro n^2 D^4)
cp = power/(ro n^3 D^5)
cq = torque/(ro n^2 D^5)
eff = jj*ct/cp

V in ft/s
n in rev/sec
D is prop diameter in ft
thrust in lbs
ro in slug/ft^3
power in ft-lb/sec
torque in ft-lb

The cart must start exceeding WS at a significant velocity and angle to generate lift in the direction to accelerate the cart.... too late, the cart has to be quite above WS for this to occur. So at best, once the cart arrives quite a bit above WS, a non DDW could generate a minor force on the pylons to help the cart along. Unfortunately for the hypothesis, once there, the cart has substantially already exceeded WS. No joy on this theory!

If need be I can look up the polars for the symmetrical airfoils we used on the pylons. I'm pretty sure they will produce a net drag in their longitudinal configuration at any reasonable angle to the wind (say up to humber's magical 40 degrees?).

 

[Jack by the hedge]

OED
4.4 Comb. auger-hole, the hole drilled by an auger.

phrases.org.uk
"Of course, 'auguring well' has nothing to do with drilling neat holes but derives from the Roman augur's prediction of a good outcome as the consequence of some portent."

 

[Llyricist]

If need be I can look up the polars for the symmetrical airfoils we used on the pylons. I'm pretty sure they will produce a net drag in their longitudinal configuration at any reasonable angle to the wind (say up to humber's magical 40 degrees?).

It'll actually get to net thrust at higher angles?

 

[windgrins]

I'm surprised you say this!

On quibble, the lift and drag of those pylons will probably never have a resultant force vector pointing forward in a relative wind coming from the front quarter.

Llyricist, this is surprising you say this. The keel on my sailboat is a symmetrical airfoil shape and I believe it indeed generates a component of its lift vector in the direction of travel generated by the sideslip which is the equivalent of an angle of attack. Otherwise my boat wouldn't sail. See Spork's "Watermelon seed" description of sail/keel interaction.

I think this would cause a problem also with aerobatic gliders when inverted.

See "The symmetry of Sailing" for a complete analysis.

However, do not mistake this in anyway as a support of Humber's argument. My previous posts still stands with regards to this!

WindGrins

 

[Llyricist]

Llyricist, this is surprising you say this. The keel on my sailboat is a symmetrical airfoil shape and I believe it indeed generates a component of its lift vector in the direction of travel generated by the sideslip which is the equivalent of an angle of attack. Otherwise my boat wouldn't sail. See Spork's "Watermelon seed" description of sail/keel interaction.

I think this would cause a problem also with aerobatic gliders when inverted.

See "The symmetry of Sailing" for a complete analysis.

However, do not mistake this in anyway as a support of Humber's argument. My previous posts still stands with regards to this!

WindGrins

Not positive, but I'm pretty sure the keel provides part of the equal and opposite drag force to the sail's thrust....

 

[humber]

I see.

I thought that, by "retreating", you meant "moving away from the air behind the propeller", which is the same as "moving forward relative to the air". But actually you meant "moving backward relative to the air behind the propeller", which is the same as "moving toward that air".

Do I have that right?

Let's knock this "retreating prop" argument over. All points of the prop are symmetrical about the hub. All points on the prop, move the same distance forwards as the hub. There is at no time any motion other than that.
Think of the prop as a solid disk and it becomes obvious that must be the case.

 

[humber]

Here is the data for the nominal prop design. As is typical, I had to make some minor changes to the planform and thickness at the root and tip to make it manufacturable (e.g. so we could fit the carbon windsurfing mast inside it). The changes should very slightly degrade these numbers - but probably still fall within the limited accuracy anyone would assign to these theoretical results:

vc=18*1.467 !vo (f/s) (this is 18 mph airspeed)
omega=120 !RPM
rho=.00226 !Density (slug/ft^3)1500 ft MSL
vsnd=1116.5 ! Speed of sound (f/s)
rt=8.0 !Tip radius (ft)
rh=0.5 !Hub radius (ft)
nblade=2 !Blade number
ao=5.73 !Lift curve slope of 2-d section
apzl= -5.0 !zero lift angle of attack (deg)
ld=67 !Section max L/D (6412 @350000 Re)
cl=0.9 !CL at max L/D
pflag=1 !1=fixed thrust 2 = fixed power
qdes=50. !design thrust (lb) or power (hp)

Design point:

Thrust= 49.9167099(lbs)
Torque= 124.131477(ft-lb)
Power= 2.83614945 (HP)
etap= 0.845000148


jj = v/nD
ct = thrust/(ro n^2 D^4)
cp = power/(ro n^3 D^5)
cq = torque/(ro n^2 D^5)
eff = jj*ct/cp

V in ft/s
n in rev/sec
D is prop diameter in ft
thrust in lbs
ro in slug/ft^3
power in ft-lb/sec
torque in ft-lb

Is that your prop, or Steve Morris'

Yes, a dump from the programme, and set at a deceptive 120rpm. Pull the other one.

The underlined figure is the power input to the prop shaft to produce that thrust.
What is missing from that load of spurious data, available to any one who can use a PC, is where that power comes from, support for a prop that can produce thrust when freewheeling, the static cases that exist when the prop is not turning at 120rpm, but barely at all, and at windspeed.

Then you can show me the mathematical support for all those claims.

Not to mention Dr Hepperle's dismissal of your naive "screwing prop" and the formula for thrust the demands input power. That last one leaves you with only ground power, as "extraction" from the air has already been refuted by three means. A table of readily available and general data, will not suffice to support your claim, in face of the author's contradiction to them.

And happily, your remark about the pylons shows you are not even close to the idea, so all will be a surprise to you.

ETA:
Thrust= 49.9167099(lbs)
vc=18*1.467 !vo (f/s) (this is 18 mph airspeed)
That is rather conveniently the value given in the NALSA submission, but happens at an airspeed of 5mph ,and an omega that is closer to 50rpm.
Anymore underhanded tricks up your sleeve?

One more. The submission has the wheel power at "400mph-pounds", which if decoded into any conventional unit, is about 1HP.

 

[windgrins]

No help for Humber either way, but let me think!

Not positive, but I'm pretty sure the keel provides part of the equal and opposite drag force to the sail's thrust....

Hmmm! Let me think about this a bit more. If I'm right, there is zero support of Humber's argument. If I'm wrong, there is even less support of Humber's argument. But I would like to give this a bit of thought. I'll report back!

Regards,

WindGrins

 

[Llyricist]

Hmmm! Let me think about this a bit more. If I'm right, there is zero support of Humber's argument. If I'm wrong, there is even less support of Humber's argument. But I would like to give this a bit of thought. I'll report back!

Regards,

WindGrins

See page 4 of "Symmetry of Sailing" for the vectors.

 

[windgrins]

Llyricist, I think you are correct!

Not positive, but I'm pretty sure the keel provides part of the equal and opposite drag force to the sail's thrust....

Llyricist, I think you are correct. Clearly a symmetrical airfoil can produce lift with a positive angle of attack. But since the airfoil is aligned with the direction of motion of the cart, Without skidding, the total force vector (lift + drag) will always be pointed behind the perpendicular to the pylon-impeding the cart in the direction of travel.

So after rethinking this, you certainly seem correct. I don't think this cart is skidding so this quibble goes to you. A very astute comment!

The only time the pylon can help the cart (although very poorly) is when there is a true tailwind. It can't help the cart past WS. So....no help from the pylons.

0-Humber
1-Llyricist
?-WindGrins

Regards,

WindGrins

 

[humber]

Hmmm! Let me think about this a bit more. If I'm right, there is zero support of Humber's argument. If I'm wrong, there is even less support of Humber's argument. But I would like to give this a bit of thought. I'll report back!

Regards,

WindGrins

You do that, whatever it is that you think you have. And guess what? The cart has no keel, and can turn on the rear axle because of the differential provided by the ratchets - should the matter of lateral thrust be at all important.

Llyricist, I think you are correct. Clearly a symmetrical airfoil can produce lift with a positive angle of attack. But since the airfoil is aligned with the direction of motion of the cart, Without skidding, the total force vector (lift + drag) will always be pointed behind the perpendicular to the pylon-impeding the cart in the direction of travel.
So after rethinking this, you certainly seem correct. I don't think this cart is skidding so this quibble goes to you. A very astute comment!

0-Humber
1-Llyricist
?-WindGrins

You also flattered Newbie over at TR with your approval, as you tried to talk your way around his argument for balloon PE.

 

[windgrins]

Huh?

The cart has no keel, and can turn on the rear axle because of the differential provided by the ratchets - should the matter of lateral thrust be at all important.

Does anyone want to explain what Humber is getting at here? It's indecipherable to me.

WindGrins

 

[humber]

Does anyone want to explain what Humber is getting at here? It's indecipherable to me.

WindGrins

Good.

 

[spork]

It'll actually get to net thrust at higher angles?

I suspect you could get a very small amount of thrust at high angles. But it depends on whether that particular airfoil becomes stalled before reaching that angle. If you look at the very broad rounded inlets for Hiller's hovering platform you can get the idea of a foil that is capable of doing that. The fairings we used for our pylons were only designed to reduce drag, so I suppose it's likely that they will stall before reaching such an angle to produce any forward thrust. Perhaps I'll check the polars. If nothing else it will give humber something to make fun of without having a clue what he's talking about.

Llyricist, this is surprising you say this. The keel on my sailboat is a symmetrical airfoil shape and I believe it indeed generates a component of its lift vector in the direction of travel generated by the sideslip which is the equivalent of an angle of attack. Otherwise my boat wouldn't sail...

The keel on your sailboat only has to provide a reasonable lift/drag ratio. It certainly doesn't have to be able to produce a lift vector whose "forward" component exceeds the drag vector's "aft" component. That's a sort of arbitrary requirement, but one that's probably not often achieved with any typical airfoil.