angle of attack ?

[willib]

if these have a 6 degree angle of attack,does that mean the tips are  0.3 + 6 degrees?

or should the tips be 0.3 degrees ? when done

i'm a little confused on this point.

http://www.otherpower.com/images/scimages/2965/angle_question.JPG

[SmoggyTurnip]

In the diagram I used 3 degrees instead of .3 degrees (just so it

would be visible)  6 degrees is the angle that the wind will appear

to approach the blade when operating at designed TSR.

I included a second diagram to show what drop and thickness

mean because I was a bit confused by these terms for a while.

Hope this helps.

SmoggyTurnip

.

[SamoaPower]

As I've pointed out previously willib, the Cl (Coefficient of Lift) and AOT alpha (Angle of Attack) are derived from the measured polar plots of a given airfoil. They are taken from the point of maximum L/D (lift to drag ratio). You enter these values into the calculator. This is why it's necessary to use a documented airfoil for good results.

Beta is the actual AOT relative to the plane of rotation. In your example, .3 degrees is correct at the tip.

Unfortunately, SmoggyTurnip's upper drawing shows a negative 3 degrees AOT. Although it can go negative at the tip for certain airfoils depending on TSR, it's usually a small positive angle.

[SamoaPower]

I need to correct a couple of things I said.

After looking at the calculator again, I see that Beta is not the actual AOT but the blade angle.

Apologies to willib and SmoggyTurnip.

[finnsawyer]

The flat side of the blade always faces the wind.  The blade is angled so that the wind hitting the flat surface will tend to push the blade in the desired direction with the rounded end of the blade moving forward.  This angle depends on both the "angle of attack" and the apparent wind.  The apparent wind depends on where you are on the blade.  Anywhere along the blade the apparent wind is the vector resultant (sorry) of the blade's speed at that point and the incident wind.  At the tip the blade speed is equal to TSR times the wind speed.  From this we get the angle of the apparent wind as the inverse tangent of the incident wind speed over the tip speed, which when you factor out the incident wind speed becomes simply the inverse tangent of 1/TSR.  For a TSR of six, for instance, you get an angle of 9.46 degrees.  You can calculate a similar angle anywhere along the blade simply by taking the blade's speed at that point equal to the tip speed times the radius at that point divided by the tip radius.  Then find the inverse tangent of the incident wind speed divided by the blade speed at that point.  For the case given, half way up the blade we get the inverse tangent of 1/3 or 18.4 degrees.  This angular dependence would be what you would use for an angle of attack of zero.  For a non zero angle of attack you simply subtract the "angle of attack" from this angle everywhere along the blade.  For an angle of attack of six degrees at the tip you get 9.46 - 6 = 3.46 degrees.  At the half radius you get 12.4 degrees.  This is the angle between the flat side of the blade and the plane in which the blade assembly lies.  That is, at right angles to the axis of the wind mill.

[willib]

GoeM thats a lot to digest , and i shall try to take it all in , in the comming days .

http://www.otherpower.com/images/scimages/2965/resultant.JPG

shouldnt you be using the chord line , instead of the flat surface? or is that what you ment?

is the apparent wind the relative wind in the .JPG ?

and what is the angle  gamma in the .jpg?

[SamoaPower]

Gamma is the AOT at max L/D. Usually 4-6 degrees.

[willib]

i asked  paradigmdesign where he got the airfoils that i used , but havnt got a reply yet . if you look in his files the shape is there.

it looks , and i thought it was a standard airfoil, donno though

[willib]

 check this out

this is now the largest transformer i own

http://www.otherpower.com/images/scimages/2965/IM002110.JPG

http://www.otherpower.com/images/scimages/2965/IM002112.JPG

http://www.otherpower.com/images/scimages/2965/IM002113.JPG

[willib]

thanks

i dont use drop , i use the actual angle.

and i dont use thickness either i use chord length, because my blades are made of airfoil sections

[finnsawyer]

I looked at the drawing.  Gamma is the angle of attack.  The sum of gamma and beta is the angle between the apparent wind and the plane of rotation, which is the angle you get for a zero angle of attack.  The amount of attack angle you introduce reduces this to the angle beta, which may become quite small or even zero.

The drawing shows the situation near the root.  As you approach the tip, the blade speed will increase until it is TSR times as large as the incident wind.  In order for the blade to work properly in that case it must be rotated clockwise from the position shown.  That is, the angle beta must get smaller.

The chord would be the proper line to use, and may be known for NACA airfoil profiles.  For a home brew blade assembly it may not be known.  The back side of the blade would certainly be known.

Here's something to ponder.  If you make your blades from a 2 by 6 (really a 1.5 by 5.5), how close can you get to the root and still have the blade operating within its minus four to plus twelve degree aerodynamic range?  I mention this because everyone says the losses at the root are negligible.  But that 2 by 6 can't give the twist demanded by the apparent wind much below half radius, at least in my example.  Once the blade is no longer aerodynamic you are just dragging a hunk of material through the air with all drag and no lift. This means power must be robbed from the rest of the blade.

[willib]

"But that 2 by 6 can't give the twist demanded by the apparent wind much below half radius, at least in my example.  Once the blade is no longer aerodynamic you are just dragging a hunk of material through the air with all drag and no lift."

i couldnt agree more!

in the first pic the blades are actually spinning , but with the flash it stops action..

[willib]

its an sg-6050

[SamoaPower]

You're on your way!

If you haven't yet, take a look at

http://www.prod.sandia.gov/cgi-bin/techlib/access-control.pl/2002/023101.pdf

They chose the SG6050 for this project. An FX airfoil performed better but they thought it was too thin for their construction technique. They confirm that the giant root section is neither necessary nor desirable. They also confirm that linear taper, which is a whole lot easier to construct, is the best compromise - less than 1% degradation in performance.

They didn't choose linear twist which I would have - it's worth ohly a few percent and again, is a lot easier to construct.

I took a quick look for the SG6050 polar plots but, didn't find them. I'm sure they're out there somewhere.

Wouldn't it be nice to hot-wire cut your blade in one swath? Please allow for spars, you need them.

Good luck.

[willib]

"Wouldn't it be nice to hot-wire cut your blade in one swath?

are you talking from root to tip in one pass?

thanks for the pdf , lots of info there.

as far as the larger set is concerned , i can still dissassemble them and hotwire a notch for glass tape no problem,

when i took the smaller set out on a windy day i was surprized by the amount that the wind was pushing the tips back from verticle, they had  no glass support whatsoever.

[SamoaPower]

Exactly!

If you use linear taper and twist, the spanwise points are straight lines so they can be cut with a hot wire if the cutter is long enough. Easy enough to make a long bow cutter.

I wouldn't suggest you use glass tape for your only spars unless you're talking about blades up to about two feet or so. I'm not sure you appreciate the bending forces on a blade in a stiff wind when you're extracting appreciable power. Think about tubular or I-beam spars. The honeycomb spars Sandia used on that project would be great if you could find them. You might also think about using shaped hardwood leading and trailing edges.

[willib]

"You might also think about using shaped hardwood leading and trailing edges."

hmmm i thought about doing that ,

did you happen to find the  plot of L/D vs angle of attack for the sg6050?

i must say i havnt lookd yet , but if you find it let me know?

[finnsawyer]

One has a lot of options in designing blades.  While maintaining the attack angle for maximum Cl/Cd may seem like the best option, as one moves toward the root the large span that results may actually give more drag than necessary.  If one allows the attack angle to increase toward 12 degrees (stall) for a good air foil design, a narrower blade profile may actually give less drag for the same lift.  That is, we should really be looking at the ratio of lift force to drag force rather than looking at the ratio of the coefficients.

[willib]

just a little update , the larger blades are now 7.5 feet dia.

and i still have one more section to make.at this point they will be at least 101 inches in dia

since i can swap out sections , i'm thinking of redoing the root , since that is the one  i started with , i can do  better now.

the hight of the origional root is 3.5" , and three  sections are 5" , three are 4 3/8 " and one is  5 3/4" , but dont tell anyone , because looking at them you would never know.

the reason for the variations is simply availablity of  foam.

the first foam i started with was 7/8" thick , then i bought the 1" thick blue foam , which sucks because its warped around the edges , it must be the extrusion process , then i bought 2" thick white insulation foam which is really 1 7/8 " thick .so three glued together is 5 3/4"

so if i make the root 5 3/4 high and the last section the same they will be 106" in dia , which is 8.83 feet..

some background : they a tsr 6 blade set

 the root chord is 21" long at 42 degrees and spans 19 degrees to 23 degrees where  the chord is also 21 inches   ,

the next section chord length goes from 21 to 18" and  spans from 23 degrees to 14 degrees,

the third section goes from 18 inches to 13  and spans from 14 to 9.5 degrees

fourth goes from 13 to 9.75 " and spans 3 degrees from 9.5 to 6.5 deg.

fifth  goes from 9.75 to to 7.3 inches  , span is 2 degrees from 6.5 to 4.5 deg

six  goes from a chord of 7.3 " to 6.0 " and spans from 4.5 to 3 degrees

seven goes from chord 6" to 5.15 inches and spans from three to 1.9 deg

eighth section chord starts 5.15" and goes down to 4.5" and spans from 1.9 degrees to 1 degree

lastly , the ninth section chord will go from 4.5" to 4" and span from 1 deg down to 0.3 degrees

thats a lot of typing , for me anyways , it started as a little  update , oh well

[willib]

the compression and  tension forces fron tip to root  are great indeed. as i saw with the smaller set the tips were bent back a good three  inches , while they were outputting 1.5A at about 4V , 6 watts into a resistive load ( an ammeter).

i think i can take care of  the tension , its the compression on the back side ,that i'm not exactly sure how to do it yet