Dear all,
I need basic step for calculating proper wormgear,slewdrive & motor for solar tracking application. with wormgear we can achieve +/- 75 degree,
660 minutes devide in 150 degree ;4.4 degree per minute.

Motor calculation as below

• efficiency= (output Power of Worm gear/ Input Power of WormGear) = Po/Pinput
  
• Output power =P0=(2*pi*Outputspeed*outputtorque)/60
  
  Output power in Kw
  Output speed in rev/min
  output torque in Nm
  
• Input power = Pinput=(2*pi*Input speed*Input torque)/60
  Input power in Kw
  Input speed in rev/min
  Input torque in Nm
  
• Net Torque= Toutput+TInput+Tholding =0

Basic details of worm gear as below:

There are lot of parameter in case of worm gear which are given below as example:

• 1 st Stage Ratio of Worm Gear: 1/10
• 2 nd Stage Ratio of Worm Gear:1/73
• Rated Output Speed: 0.078rpm
• Rated Output torque: 910 Nm
• output torque :1300Nm
• holding torque :10.4Knmt
• axila load : 133KN
• radial load :53KN
• o/p speed : 56.9rpm

These are basic equation followed usually

In order to drive 250Kw plant ; assume 15kw per row I have 16 rows with 1 column.

Worm gear has to Drive 16 rows with coupling with single column item

each panel weighs 20Kg ; 60 panel per row-> 60*20kg=1200Kg
Each row structural weight around = 700Kg

Net weight alone = 1900Kg per Row

so load on worm gear might be 1900*16=30400

• Basic infomation collected while selecting worm gear as below
• if higher torque; higher the gear ratio; speed @ output will be low & vice versa
• if Input gear has low teeth than output speed will be lower @ output.
• larger the Movement of interia the greater torque requied to accelerate & deaccelerate

My question are as below

1. based on above value how can calculate worm gear & motor spect
2. how to determine radial load & axial load .
3. In order to run 4.4 degree per minute. what must be gear ratio of worm gear choosen with their speed.
4. How should i know rpm @ which output of wormgear must be run.

if someone can give exact calculation part here . it will be more helpful. I know we should calculate from output to input.


https://www.anaheimautomation.com/products/stepper/stepper-motor-item.php?sID=20&pt=i&tID=75&cID=19

http://www.sunflowerenergyinc.com/Slewing_Drive_-_032511_-_REV.03.pdf

 

The wind load is at least 100X larger than any mass related load, inertia, or bearing friction. I know this from my antenna rotator...

 

anyone can help me to find proper relationship with worm gear & motor

 

consult a machinist handbook.
if your worm has 2 threads per inch and your driven gear has 40 teeth. your gear will revolve through one complete revolution for every 20 turns of the worm. Make the gear have 50 teeth and it will take 25 revs of the worm to drivr the gear through one revolution.
extremely ODD ratios are hard to work out and will be impossible to find in off the shelf products.
I suggest find a suitably slow ratio of whole number gear reduction taking into account the commonly found gears and then controlling the motor speed with a suitable PWM drive controller to get your exact speed.

 

I am intrested to know how the calculation done here. Assume my track time is 660 min , with document said that with worm gear we can acheive +/- 75 degree i.e total 150 degree . 4.4 degree for minute. Inorder to match that @ what speed stepper motor must be run.for example these specification for slew drive.

1 st Stage Ratio of Worm Gear 1/10
2 nd Stage Ratio of Worm Gear 1/73
Rated Output Speed 0.078 rpm
Rated Output Toque 910 N.m
Output Torque (Max) 1300 N.m
Tilting Movement Torque (Max) 13.5 kN.m
GEARBOX PERFORMANCE DATA Holding Torque 10.4 kN.m
Output Torque 105 kg.cm Axial Load (Max) 133 kN
Output Speed 56.9 rpm Radial Load (Max) 53 kN
Voltage 24 VDC Tracking Precision 0.1 – 0.14”
Rated Current 3 A Hall Sensor Sensor

 

Is this some kind of assignment question?

 

you have a gear reduction of 730:1
You wish to rotate through .4166666..... (150/360) of a whole circle.
You have 11 hours to perform this partial rotation.
Interestingly...11/24 is .45833333...
You want a speed that will perform a full rotation in about 27.5 hours.

27.5 x 60 = 1650 min.
gear reduction is 730
if you can't get an approx RPM at this point then maybe someone else should install the panel rotation system.

 

You want a speed that will perform a full rotation in about 27.5 hours.

27.5 x 60 = 1650 min.
gear reduction is 730
if you can't get an approx RPM at this point then maybe someone else should install the panel rotation system.


i could not able to get this point. How you get 27.5 hours??
Tracker will run from morning 9AM to 6PM .i.e total 9 hours .
And another point In order to run this how much input speed of motor i should run . What is Input power required ??

 

Here's a starter:
1) Express the tracking speed TS in RPM,
2) Choose a realistic motor speed MS in RPM (research needed),
3) Calculate MS/TS to give the required gear reduction ratio,
4) Research what gears are available to give that reduction ratio in 2, 3, maybe 4 stages,
5) Calculate the torque values for that ratio, bearing in mind Mike's comments in post #2,
6) Check if gears and motor are available which can withstand/provide the torque required,
7) If not available, repeat steps 2-7 for other speeds.

 

Yes that only i asked for .
1) Express the tracking speed TS in RPM,
How can i calculate tracking speed in rpm?? if i am going to run for 9-10 tracking hour as kemit say 10/24=0.4166 or 150/360=0.4166 .

2) Choose a realistic motor speed MS in RPM (research needed),
In order to calculate the speed we must know answer for first on . depend on gear ration & speed ratio reation i.e g1/g2=n1/n2 we can determenine speed required at input of slew drive then we calculate output power of motor based on relations.

3) Calculate MS/TS to give the required gear reduction ratio,

4) Research what gears are available to give that reduction ratio in 2, 3, maybe 4 stages,
didn't get this point
5) Calculate the torque values for that ratio, bearing in mind Mike's comments in post #2,
6) Check if gears and motor are available which can withstand/provide the torque required,

7) If not available, repeat steps 2-7 for other speeds.

I need simple ample explaining.Here i have attached motor & drive specification. if i choose the above drive & motor . how can calcuated above all your stage . with related explanation clear my doubt. Like kermit #7 commented
motor specification:https://www.anaheimautomation.com/products/stepper/stepper-motor-item.php?sID=20&pt=i&tID=75&cID=19
slew drive specification:http://www.sunflowerenergyinc.com/Slewing_Drive_-_032511_-_REV.03.pdf

 

first you wanted 11 hours, now you want 9 hours.

How about you change to a drive on demand system?

put two photocells, one on either side of a panel. place them a few inches back so that a shadow will fall on one as the sun moves. have the motor come on and drive the panel array till both photocells are illuminated again. same result as solar tracking with constant drive.

because I know you will ask: you use two cells to detect a difference of light level. as clouds or night time occur both photocells will be equally illuminated and circuitry will not see a 'differnce' of light levels needing correction

 

I'll do step 1 for you.
From your post #1 figures, slew rate (tracking speed) = 4.4º/min = 4.4/360 RPM = 0.012 RPM.

In step 2 I didn't suggest calculating the motor speed.
Re step 4, it's all very well calculating some theoretical gear ratio, but you need to do a reality check and find out what gearing is actually available to you. Your figures gave example ratios for two stages, but methinks you may well need three or more stages in practice.
Since this is your assignment, I leave it to you to tackle the other steps .
The stepper motor series you linked to has a max torque of 320 oz.in. Convert that to N.m and see how that figure compares with the output torque needed to slew your solar array. You may be surprised!

 

And you never dealt with the wind loading issue. The gear-train and motor torque rating needs to be 10X to 100X more than just simply moving the mass of the panels...

 

The geometry of the solar array is also important. If, for example, the weight and wind-exposed surface area of each row of panels are perfectly balanced about the axis of rotation, then less torque will be needed to rotate it than if it were unbalanced.

 

as far as holding torque on the motor, it should not be a big deal. Worm drive is extremely resistant to counter torques when the drive is not turning.

 

first you wanted 11 hours, now you want 9 hours.

How about you change to a drive on demand system?

put two photocells, one on either side of a panel. place them a few inches back so that a shadow will fall on one as the sun moves. have the motor come on and drive the panel array till both photocells are illuminated again. same result as solar tracking with constant drive.

because I know you will ask: you use two cells to detect a difference of light level. as clouds or night time occur both photocells will be equally illuminated and circuitry will not see a 'differnce' of light levels needing correction

I am using inclinometer as feed back sensor. I am not using irridiance sensor. So you can use purchase inclinometer 0~180

 

Is this for a practical application or is it a purely theoretical exercise?

 

Tilt sensor or photocell. your choice.

Some questions for you. If you want to track the sun and you want to have gear reduction to slow the movement, how are you planning on returning the array to the starting point for the next day? Do you have a gear bypass to allow a quick return?
What kind of information will you need from your tilt sensor, and how will you use it to provide control?

I still say you need to choose from commercially available gears and motors, then use a motor controller to dial in the desired speed.

 

Most solar trackers are not constantly 'on'. Turns maybe once an hour. Would be a waste of electricity to be on all the time. Get the motor first, then go from there. The incline motors from treadmills work well if the panels are not too heavy.

 

...If you want to track the sun and you want to have gear reduction to slow the movement, how are you planning on returning the array to the starting point for the next day?

You have all night to slew the panels into position for the next morning...

And in picking the motor and gear train, the biggest force you need to overcome is wind (not gravity)