somebody check my caculations please

[gww]

I have only a volt meter.  I do almost all my figuring from 60 rpm because it is easy to count seconds.
If I wanted a good 12v turbine with 8 foot blades at 7 tsr with cut in at seven mph (3 ms) my figures are;
1.  Blade rpm at seven mph is 167 rpm.
2.   167 rpm divided by 3 is 55.6 rpm.
3.  14v divided by 3 is 4.6 volts
So a decent 12v tubine with these blades would probly have between 4.5v to 5v at 60 rpm?
If this turbine had say 21 gage wire coils rather then 12 gage wire coils.  would I want a little higher voltage then this at 60rpm to help contend with coil heating? Is any or all of my assumptions wrong?  I thank any who are willing to help with this.
Thanks
gww

9 foot blades at 7mph 148 rpm.
148 divided by 2.5 = 59 rpm
14v divided by 2.5 = 5.6v at 60rpm
   gww

[Frank S]

I have only a volt meter.  I do almost all my figuring from 60 rpm because it is easy to count seconds.
If I wanted a good 12v turbine with 8 foot blades at 7 tsr with cut in at seven mph (3 ms) my figures are;
1.  Blade rpm at seven mph is 167 rpm.
2.   167 rpm divided by 3 is 55.6 rpm.
3.  14v divided by 3 is 4.6 volts
So a decent 12v tubine with these blades would probly have between 4.5v to 5v at 60 rpm?
If this turbine had say 21 gage wire coils rather then 12 gage wire coils.  would I want a little higher voltage then this at 60rpm to help contend with coil heating? Is any or all of my assumptions wrong?  I thank any who are willing to help with this.
Thanks
gww

9 foot blades at 7mph 148 rpm.
148 divided by 2.5 = 59 rpm
14v divided by 2.5 = 5.6v at 60rpm
   gww

Ok maybe it is because I have been up for 36 hours and am not reading correctly
 but I'm confused first off you are saying that your blades rotate 167 RPMs in 3 ms wind. How did you come up with this figure? What is the chord area of your blades what is the air foil profile and or pitch angle
 whether a blade set is 3 feet in diameter of 50 feet in diameter RPM is the revolutions the shaft turns in 1 minute
 You state you are dividing by 3 do you have a speed reducer gear box with a 3 to 1 ratio or  are you counting the individual blade tips as they go by then dividing by 3 to come up with the RPMs
 Like I say I'm confused

[gww]

Frank
I used hugh piggot blade desine and his math.  "rpm=windspeed x tsr x 60/circumference". 
wind speed 7mph is 3miles per second.  I then devided the answer by a number that would give me close to 60rpm.  I then took my target cut in voltage and devided it by the same number that I used to reduce the rpm to 60.  It is direct drive.  60 rpms can be counted by seconds so, with no measuring device this is a posible way to (reliably?) check out put.  hopefully I can get the altinator to match the blades with this prosess.  14v at 167rpm = cut in at wind speed 7mph. I also wound with fairly thin wire for a 12v machine and need to know if I have to compensate for this.  thanks for the question and If I dont make sence please ask and ill try harder.  If you know the answers please help me if you can.
Thanks
gww

PS  36 hours is a long time, hope I was self imposed.

[bob g]

"wind speed 7mph is 3miles per second"
 
you might want to check your  math with this one!  3 miles per second is something on the order of 3x60 seconds x 60minutes=10,800mph?

lets see what we have here

"rpm=windspeed x tsr x 60/circumference".  (assuming this is correct)

lets take the windspeed in ft/min

10mph would work out to 10x 5280 = 52800ft/hr
52800 / 60 min = 880ft/min

so... rpm = 880 x tsr x (60 x circumference)

(looks like something is wrong with your equation don't you think?)

rpm = 880 x tsr 5 = 4400  (60 x circumference)

rpm = 4400 x (60 x circumference)

lets solve for circumference and it needs to be in feet.

so for a 10 ft rotor we have 10 x 3.14159 = 31.42ft

now

rpm = 4400 x (60 x 31.42) = an insane rpm!

what we should have is the following

rpm = windspeed in ft/min * tsr / rotor/blade circumference  (i think, so lets see)

from above a 10 mph windspeed works out to 880ft/min
so
rpm = 880 * tsr (lets use 5) = 4400 /  circumference

rpm = 4400/  circumference ( lets use a 10ft rotor diameter)

circumference = 10ft * 3.14 (pi) = 31.4 ft

therefore
rpm = 4400/ 31.4 = 140rpm
so lets cross check

a 10 ft rotor turning 140
a 10 ft rotor is 31.4 ft in circumference
31.4 * 140= 4396 ft/min
4396 /5280 = .8325 mile/min
.8325 x 60 = 49.95mph
49.95 / tsr 5= 9.99 mph

so that works out, i rounded the numbers off so that is why we don't get exactly 10mph wind in the cross check.

i think i have this right, someone will correct me if i am wrong

fwiw
bob g

[Flux]

You look to be close. tsr7 on an 8ft prop came out by my rough approximations to be 180rpm, your 170rpm is probably right and it seems a good cut in speed for Hugh's 8ft prop.

You would be looking  at about 1/3 of the battery voltage at 60 rpm as a dc figure after the rectifier so your 4.5v is ok.

The last part of the argument goes a bit wrong. If you wind with thinner wire the stator heating problem becomes worse and adding more turns will only add even more resistance and make it worse. The only benefit you will get with more turns is that it will stall harder and the reduced current output may keep the stator heating down while it is hard stalled but if it gets away then any extra resistance is working against you.

A better approach would be to use thicker wire and use all the winding space, just use the minimum turns to get the cut in and have the lowest resistance but if you have the stator then use it, increase the air gap to get the cut in right ( 4.5v at 60 rpm) and make sure the furling is set light enough to keep the maximum stator current within limits for your thinner wire. With the cut in right and the thinner wire I wouldn't expect it to stall so you will have to rely on furling for protection.

Flux

[gww]

Bob g
Maby its not miles per second. Meters per second?  The formula I used was 1 m/s = 2.2 mph
Thank you for yor time.
gww

Flux
This is the stator that was already built that I put neos on instead of the original ferrits.
I was trying to take your original exellent advice from my other post and make sure I understood what the goal was.
From your responce here it seems that I understand the issue now well enough to move forward in adressing it.  I shouldnt have to bother this board with this perticular problim any more as your answer makes everything clear and I now understand completly what I have to do. I cant thank you enough for having the patients to help me get to this understanding.
sincerly
gww

[Frank S]

There is a handy Wind turbine blade calculator at
http://warlock.com.au/tools/bladecalc.php



Wind Turbine Blade Calculator
 
 
   3  Number of Blades    SI

 
   7 TSR   
 
 .4  Blade Efficiency    Metric

 
 4  Blade Radius (ft)   
 
 7  Wind Speed (mph)    Imperial

    
 

 
 
Power   34.9 Watts
Rotational Speed   
171.6 rpm
Torque   1.94 Nm

Radius (ft)   Chord (in)   Beta (deg.)
0.4   21.516   37.6
0.8   10.752   19.5
1.2   7.176   11.6
1.6   5.376   7.4
2   4.308   4.8
2.4   3.588   3
2.8   3.072   1.7
3.2   2.688   0.8
3.6   2.388   0
4   2.148   -0.6

[gww]

Frank
Thank you for the link. 
gww

[gww]

I lied.
 
I said I understood and wouln't have to bother this board with this turbine issue again.

Flux's answer caused me to think That I had one more thing I need to know.

  I need figure out what wind speed to furl at that does not exceed 8 or 9 amps for 21 gage wound coils.  If anyone knows What that wind speed would be with the eight foot blades that would be great. If I have to figure it out thats fine too.

  Am I understanding correctly;  wire heat limit would be 8 or 9 amps and this is what I have to furl at?

 Is this what I have to do?

sorry
gww

[gww]

Is 8 or nine running amps what I'm looking for and Is there a chart somewhere showing magnet wire gage amp capabilities?
gww

[Frank S]

Is 8 or nine running amps what I'm looking for and Is there a chart somewhere showing magnet wire gage amp capabilities?
gww

try this link
 then Rt click the page save as or print, then what ever format you what to save it in
http://amasci.com/tesla/wire1.html

[gww]

Frank s
Its a shame when you get what you ask for but are too slow witted to use it.  Could you help me find a simple rule of thumb to use this chart as it is above my head. Perhaps like my stator has six pounds of copper so amps are, or when I use an ohm meter I believe when I tested my last coils with tester on the lowest ohm test 200- I got  between 40 and 39.  Anyway thanks for the link
gww