Math formula

[JJPage]

I have a 50ft Rhon tower with a hinged base for my base radios. Amateur Radio Supply in Seattle has some decent winches that may work for you, both hand and electric.

 

[BIGNASTY]

JJ and others

Thanks for the info but I think im going to stick to the plan at the start of the post. On our Wind gen we have a tower that looks very much like some of the ones that you suggested and I have at times worried about the W Gen. Talked to the FIL/Son and they said if we need to get a bucket truck to hook things up do it. My only concern was the counter weight im not 20 years old and I dont bounce well anymore getting up on ladders with that kind of weight is NOT GOING TO HAPPEN with me. Thanks again for all the info in the post it is appreciated I would still be working on 2+2= 1200 :-laf



BIG

 

[GaryOwen]

Looks to me the weight to balance would be calculated as follows:



Load Force = (450x20)/12



or 750 pounds



Maybe I'm missing something?

 

[rMetzger]

Gary's got the math right for balancing a heavy load on a short lever and a lighter load on a longer lever. The big question in my mind is: Will the tower support 1300lb with some of that weight being a spinning windmill? It'd be a mess if, after all the multiplying, dividing, adding and subtracting, a shift of wind gust flexes any of it beyond the point-of-no-return and collapses the whole thing.



another consideration would be this: when you're on the way "up" or "down" you're going to pass thru a zone where the upper "arm" of the tower is level-to-the-ground, and there'll be maximum stress on the place where the lower part of the tower is the bottom of a letter "T", and the upper arm might buckle. You might do well, during those times, to get a guide-pole, shaped somewhat like a tall letter "Y", that you can use to stabilize the expensive end, such that ALL the pressure isn't just on that one little pivot-point.

 

[BIGNASTY]

I thought the same thing about all the weight there are several people up here that are off gird like us, most have BIGGER wind gens than us but have smaller solar systems and NO backup gens to speak of they have your little Honda's and Yamaha's but that is just for emergency for them. The wind gen tower in the first post when you look at it when its in the in-use position called flying in wind gen terms the bulk of the weight would be pushing DOWN on the main part of the stand. Their are far bigger gens on far less mounts up here, and like my Son noted that I WAY OVER BUILD EVERYTHING!!! The pad for the mount was put in last fall when you could still dig in the ground the cold has it like concrete now same with the guy-wire supports dead-men there is more concrete in this install than any I have seen the cable to be used is also bigger. And if it gets that bad bring it down I take ours down when it gets TO WINDY (sounds funny for a wind gen TO WINDY) but it does happen. I dont think that twisting is going to be a problem.



Even the commercial wind farms turn the gen's away from the wind at times in So Ca at White Water pass they stop them ALLOT!!! because of high winds.

 

[fest3er]

Looking at this the triangle is the point to where it is attached to the upright poles cemented into the ground the (the pivot point)



The Effort force would be the wind generator including the weight of the pole from the pivot point to the end 450lbs the distance for pivot point to end of wind gen side of lever would be 20 ft (dont know if you need that)



The land force would be the attachment point for the winch cable to pull down and that would in turn raise the wind gen. The distance of the lever from the pivot point to the land force end where the counter balance weight will be is 12ft.



How much weight would be needed on the LAND FORCE to make it easy to let down the wind gen without being out of control.

This math may be far more involved than you might want. But it'll get you there.



It's simple torque or work (force through distance). For a uniform weightless beam, the left side must equal the right side. In your case: 450# x 20ft = CWT x 12ft ==> 450x20/12. The counterweight must be 750# to be exact.



The 20/12 ratio should stand for any rotation of the beam, but it's good to verify. Suppose you've raised it to 75°. Is the counter weight still correct? At 75°, the effective horizontal lengths of the beam segments are 20 x cos(75) and 12 x cos(75). The two cosines cancel out (= 1) and the ratio remains the same: 20/12.



If the weight of your pole is uniformly distributed (as in a typical i-beam segment), get its total weight, then multiply by 20/32 to get the long segment's weight, and by 12/32 to get the short segment's weight. Suppose your pole weighs 160#. The weights will be 160 x 20/32 and 160 x 12/32, or 100# and 60#.



Divide the weights by 2 (it's a calculus thing), and redo the equation:



(30#+CWT) x 12ft = (50#+450#) x 20ft, or



CWT = ((500#) x 20ft/12ft) - 30#.



CWT= 803⅓ pounds to be exactly balanced, for this example. An 810# counterweight would let you move the whole thing with one hand and ensure it can't 'fall' by itself. But remember the total *mass* of the device; once it starts rotating, it ain't gonna want to stop. A judiciously placed shock absorber would dampen its movement and decelerate it as it reaches horizontal and vertical. And let you exactly counterbalance the generator.



A final note on the equation. The distance assumes the weight is uniformly distributed at each extremity. If your counterweight is going to run from 10' to 12', then you must compute using an 11' length; that is, on average, the weight will sit at 11'. If your generator is not 'balanced' (maybe balanced when upright but unbalanced when laid over), you'll have to take that imbalance into consideration; it could put 400# further out the beam when horizontal, requiring more counterweight, which will be even more weight keeping the thing upright (and making it harder to tilt down).



And the generic equation (for beams with uniformly distributed weight):



(loadLeftWeight + beamLeftWeight/2) x leftDistanceToFulcrum =

(loadRightWeight + beamRightWeight/2) x rightDistanceToFulcrum



I'm still not certain of this simple beam weight calculation. It may be more complicated.

 

[BIGNASTY]

If the wind shifts when its working the whole wind gen turns to face the wind because of the tail when these get going there is not much vibration they are balanced pretty well I have never taken a reading on the blade RPM but when the wind is up they are MOVING!!!! Had a small bird fly into ours one time not much left just a few feathers :-laf



This may be miss leading let me find a picture and show you how its mounted on the pole

 

[BIGNASTY]

This math may be far more involved than you might want. But it'll get you there.



It's simple torque or work (force through distance). For a uniform weightless beam, the left side must equal the right side. In your case: 450# x 20ft = CWT x 12ft ==> 450x20/12. The counterweight must be 750# to be exact.



The 20/12 ratio should stand for any rotation of the beam, but it's good to verify. Suppose you've raised it to 75°. Is the counter weight still correct? At 75°, the effective horizontal lengths of the beam segments are 20 x cos(75) and 12 x cos(75). The two cosines cancel out (= 1) and the ratio remains the same: 20/12.



If the weight of your pole is uniformly distributed (as in a typical i-beam segment), get its total weight, then multiply by 20/32 to get the long segment's weight, and by 12/32 to get the short segment's weight. Suppose your pole weighs 160#. The weights will be 160 x 20/32 and 160 x 12/32, or 100# and 60#.



Divide the weights by 2 (it's a calculus thing), and redo the equation:



(30#+CWT) x 12ft = (50#+450#) x 20ft, or



CWT = ((500#) x 20ft/12ft) - 30#.



CWT= 803⅓ pounds to be exactly balanced, for this example. An 810# counterweight would let you move the whole thing with one hand and ensure it can't 'fall' by itself. But remember the total *mass* of the device; once it starts rotating, it ain't gonna want to stop. A judiciously placed shock absorber would dampen its movement and decelerate it as it reaches horizontal and vertical. And let you exactly counterbalance the generator.



A final note on the equation. The distance assumes the weight is uniformly distributed at each extremity. If your counterweight is going to run from 10' to 12', then you must compute using an 11' length; that is, on average, the weight will sit at 11'. If your generator is not 'balanced' (maybe balanced when upright but unbalanced when laid over), you'll have to take that imbalance into consideration; it could put 400# further out the beam when horizontal, requiring more counterweight, which will be even more weight keeping the thing upright (and making it harder to tilt down).



And the generic equation (for beams with uniformly distributed weight):



(loadLeftWeight + beamLeftWeight/2) x leftDistanceToFulcrum =

(loadRightWeight + beamRightWeight/2) x rightDistanceToFulcrum



I'm still not certain of this simple beam weight calculation. It may be more complicated.

WOW!!!!! Im going to make a copy of this and do some reading



I could not find a picture of the mount its basically a plate that mounts or welds to the end of the pole and goes into the mount of the generator bearings on both sides at the mount and in the generator mount its all covered with plastic is the reason I cant get a pic of it. Its a pretty heavy duty set up far better than the one I have and in the picture on a maint stand for its yearly maint. The other pic is of the little one just built for the barn to run 12v RV water heater probes to heat the water for the animals so it wont freeze it also runs a single light (I milk and clean at 3 am) and my radio for some tunes to work by and charges the 2 deep cycle 12v batts.



When FIL and Son went to get this big wind gen I took some of the drawings of the pole mount (I did pretty well in Drafting class) the salesman asked what we needed for a tower I pulled out my plans and 3 guys asked your doing this? I thought OH NO They all said its an over build but it will take a bigger wind gen than what they are buying. I will try to remember to bring my camera when I go over to the site again.



My Two wind gens pictured below are toys in comparison to what they bought but they are going to run the sons whole she-bang on just it.



IT WAS SAID AWHILE BACK BY SOMEONE ON THIS SITE THAT THE SOLAR AND WIND GENS ARE JUST TOY'S THAT MIGHT BE!!!!! BUT I HAVE YET TO PAY A ELECTRIC BILL IN ALMOST 4 YEARS THAT'S !!!!!!SOME TOY:-laf



BIG

 

[Mikey-KE7LBB]

I just noticed your post and photo of the exact same wind turbine that the local school has...



I see it spinning nearly every day... Even when all of THE BIG ones have been shut down due to high winds!

 

[BIGNASTY]

I just noticed your post and photo of the exact same wind turbine that the local school has...



I see it spinning nearly every day... Even when all of THE BIG ones have been shut down due to high winds!



View attachment 83863

Its a pretty popular model and has been around for a bunch of years. I dont take mine down because the wind will hurt the generator its self its the tower that bothers me. If this works out I will build one for ours. If its the same one the blades will fold and stop the generator if the wind gets real bad. Ive seen ours going so fast that the blade looked like it had stopped the wife came running in the gen stopped!!!! We have a meter in the house I looked at it and said its charging so its not stopped its an illusion.



Last night the problem just kept picking at me and correct me if im wrong but fest3er if I were to follow your math equation that would make BOTH ENDS OF THE LEVER THE SAME!!! If this is true that is not what is needed. In essence if they were the same weight at both ends you could stop it halfway down and it would stay level. The reason for the counter weight was to AID IN THE LOWERING of the wind gen. Once it was released from operating position and kicked off center on the stand if the weight were the same at BOTH ends it wouldn't want to drop the gen. So there has to be a difference to make it come down and the counter weight makes it a more controlled procedure. I was out at the Hwy today having coffee with the state road crew talking of the concerns and one of them said to ask a couple down the road in the next town about their's. I feel kind of weird doing that but he made a call and the couple had us down. His wind machine is about the same as the one we are dealing with and he said that his tower although being a little different takes about 1/2 the weight of the gen he said to much counter weight and it didn't roll up the cable well and a few times left him in a bad situation half way down. I KNOW THAT THE THOUGHT OF THE ATV WINCH IS NOT!!!! going to work he took his down and then back up and its quite a strain on the winch and he had a smaller off road Warn so bigger is better in that area. On the going up and down it seemed that there was only a spot that it really made it work just before thru just after half way up other than that you could hear the difference in the winch. Hope I can sleep tonite things like this BUG ME TO NO END the wife said it a compulsion disorder OK SO WHAT!!!:-laf

 

[fest3er]

Last night the problem just kept picking at me and correct me if im wrong but fest3er if I were to follow your math equation that would make BOTH ENDS OF THE LEVER THE SAME!!! If this is true that is not what is needed. In essence if they were the same weight at both ends you could stop it halfway down and it would stay level. The reason for the counter weight was to AID IN THE LOWERING of the wind gen. Once it was released from operating position and kicked off center on the stand if the weight were the same at BOTH ends it wouldn't want to drop the gen. So there has to be a difference to make it come down and the counter weight makes it a more controlled procedure. ...

Yes. But no. What it takes is more mass at the upper end to make it want to fall by itself. Or more force at the lower end to move it at all.



Here are the three possibilities so you can picture it in operation.

1. You could use less counterweight. This would make it top heavy and want to fall by itself. You would need a winch or other belaying mechanism to control the descent and to raise it. And you'd have to pay close attention to the cable's condition; if it breaks while raising or lowering, it all comes crashing down.
   
2. You could use more counterweight. This would make it bottom heavy and require hydraulics or a winch to pull the bottom to the side and upward. You would have to block the counterweight to prevent it from raising the mast by itself. But you'd never have to worry about it falling down if the cable broke or the locking pins came loose. In medieval times/dark ages, the extreme version of this was called a trebuchet. Could be useful if you wanted to get rid of trash without hauling it overland.
   
3. You could exactly counterbalance it. You could then start it moving giving it a shove; momentum would carry it the rest of the way. But you'd have the devil's own time stopping it. So you'd have to use a 'double' winch that both pulls on the mast above the pivot and belays the bottom end at the same time when lowering it. When raising, the bottom cable would provide the motion and the upper cable would belay. Or you could use a hefty hydraulic dampener that works only when extending. Positioned right, you give the mast a shove, momentum carries it through, and the dampener slows and stops it at horizontal and vertical; some spring relief might be called for in case you shove it with excess vigor.
   

People usually opt for #1 because it is more conventional and doesn't require a mast to support 1400# (at the pivot; include a safety factor of 3 or 4) when horizontal.



What would you need a winch for? You have mules. Affix a pulley at the bottom a foot or two in 'front' of the mast. Hook a cable from the bottom of the mast through the pulley to the mule. She'll walk it right up. Or down. As she'd be pulling upward-ish (from the pulley), she'd have extra traction. Or use the mule to reduce your manual effort. She adds 400# lifting force, you add 100#; that leaves a 300# counterweight.



Mules aside, I'd prefer to use a standard 'tower' crank. Dad's got a self-supporting, 3-section, 55' crank-up/tilt-over ham tower, maybe 25' sections; the truss probably weighs around 600#. Upward uses the typical ratchet dog; the crank handle never moves more than an inch or so before the dog catches. Downward uses friction; the dog is never released. Cranking releases the clutch a bit, letting the cable reel turn and re-tighten the clutch. (In other words, it's self-stopping. ) Cranking up is a light workout; down a little less so. These cranks're used on antenna towers and professional stage lighting/truss rigs. (I'm sure there's a name for that kind of crank, but I've never been interested enough to find out. ) If rock concert roadies can use them without injury, I'm sure you'd have no trouble.

 

[BIGNASTY]

Have to give it all a think ive got until next spring before the install. Have some forum members from ID coming over in a week or two to take a look see. Was also thinking of making the arm of the lever without the wind gen longer just have to think of a different locking system


BUT LETS LEAVE MY MULES OUT OF THIS :-laf

 

[cojhl2]

If it were me I would balance the beam exactly and use a positive system of raising and lowering, much like an elevator. This could be done with equal diameter drums at the pivot point and at the crank end.