On Tue, 22 Dec 2015 15:43:01 -0800, Jessica Shaw wrote:> Hi, > > I am using an accelerometer to calculate tilt. Its single axis control. > Why 15 degrees an hour?Well, lessee. You're tracking the sun, and a day is 24 hours long, and the sun scribes a circle in the sky once per day, and a day is 24 hours long, and a circle has 360 degrees, and 360 / 24 = 15 and that ain't no coincidence at all.> The motor takes about 28 second to move 15 degrees. The motor takes > about 2A at 12Volts. I am trying to figure out how much current it is > drawing in 28 seconds. > > Let's say I want the motor to move 15 degrees after every hour. If it > moves from 90 to -90 then the travel is 180 degrees which means that > motor will move 12 times in a day and then after Sun set it will travel > back ( 3 minutes) to its initial position facing East waiting for the > Sun rise again. So, total number of times motor moved = 13. > > I want to calculate that how much current it will use from the battery ( > 12V, 2AH) each day plus how long ( how many days) my battery will last > without charging. How can I make these calculations?How can you make these calculations? By getting the rest of the information you need, as has been pointed out to you, repeatedly. Asking the question again and again won't make the information magically appear -- you need to actually Do Some Work to dig it up. -- www.wescottdesign.com
Motor Control
Started by ●December 18, 2015
Reply by ●December 22, 20152015-12-22
Reply by ●December 22, 20152015-12-22
I think you did not read Richard Damon's comments. Can you confirm my following calculations 56 Amp - sec? This is Coulomb. Battery capacity is 4AH. 3600 secs -----> 4A 28 secs -----> 31mA. So, the current draw on the battery for 28 secs will be 31mA. If the motor moves 13 times in a day then 13x 28 = 364 secs in a day. 364 secs ------> 400mA per day. My battery will last for 10days.
Reply by ●December 23, 20152015-12-23
On 23/12/2015 1:53 PM, Jessica Shaw wrote:> I think you did not read Richard Damon's comments. Can you confirm my following calculations > > 56 Amp - sec? This is Coulomb. > > Battery capacity is 4AH. > > 3600 secs -----> 4A > > 28 secs -----> 31mA. So, the current draw on the battery for 28 secs will be 31mA.For those 28 sec it will draw 2A as you mentioned previously. Averaged over an hour this is: 28/3600 * 2 ~= 0.015Ah.> If the motor moves 13 times in a day then > > 13x 28 = 364 secs in a day.13 x 0.015Ah ~= 0.195Ah per day> 364 secs ------> 400mA per day. > > My battery will last for 10days.4Ah / 0.195Ah ~= 20 days Possible reasons why you won't get 20 days: Are you powering some electronic control system as well? Add it to the Ah used per day. Example, if it draws 100mA average then that's a 0.1Ah drain. Check the discharge curve of your battery, they don't supply constant current at constant voltage for the entire period of charge. HTH -- Cheers, Chris.
Reply by ●December 23, 20152015-12-23
On 12/22/2015 7:36 PM, Jessica Shaw wrote:> Hi, > > 56 Amp - sec? This is Coulomb. > > Battery capacity is 2AH. > > 3600 secs -----> 2A > > 28 secs -----> 150mA. So, the current draw on the battery for 28 secs will be 150mA. > > If the motor moves 13 times in a day then > > 13x 28 = 364 secs in a day. > > 364 secs ------> 200mA per day. > > My battery will last for 10 days. Am I right?I don't exactly follow your math. Battery is 4 AH. Each move is 56/3600 AH = 0.0156 AH 4/0.0156 = 257 moves. 13 moves per day = 257/13 = just under 20 days. So yes, I guess your math worked out ok. -- Rick
Reply by ●December 23, 20152015-12-23
On 12/22/15 7:36 PM, Jessica Shaw wrote:> Hi, > > 56 Amp - sec? This is Coulomb.Yes, batteries typically have a limited number of electrons they can deliver, based on chemical reaction progressing. This count of electrons can be done in a number of units, be it Coulomb (aka Amp-Second), or Amp-Hours (often used as operational life is normally in hours, so more reasonable numbers).> > Battery capacity is 2AH. > > 3600 secs -----> 2AThis is a nonsense expression. You could say that 2 AH / 2A = 1 hour = 3600 seconds> > 28 secs -----> 150mA. So, the current draw on the battery for 28 secs will be 150mA.No, the current draw is 2A for 28 seconds. I have no idea where these numbers are coming from.> > If the motor moves 13 times in a day then > > 13x 28 = 364 secs in a day. > > 364 secs ------> 200mA per day. > > My battery will last for 10 days. Am I right? >as above, 2 AH / 2A = 1 Hour active run time. At 28 seconds per move the is 3600 seconds / 28 Seconds/move = 128 moves. 13 moves a day = about 10 days. (note, this may be optimistic, as the return move is likely longer. As others have pointed out, you also need to include the fact that your controller is consuming power too, which needs to be considered. Also, as you drain the battery, it generates less voltage, which says the load (motor) will likely draw less current, which might sound good, but it also won't push as hard and take longer to make the move. Also, a major factor is how that 2A draw for 28 seconds was determined. I actually strongly suspect that the draw will NOT be constant over the full move. You will normally get a large surge when you first start the moter, and the current will drop as the motor gets moving.
Reply by ●December 23, 20152015-12-23
Il giorno mercoled� 23 dicembre 2015 01:36:14 UTC+1, Jessica Shaw ha scritto:> Hi, > > 56 Amp - sec? This is Coulomb. > > Battery capacity is 2AH. > > 3600 secs -----> 2A > > 28 secs -----> 150mA. So, the current draw on the battery for 28 secs will be 150mA.wrong. It would be 150 mAH. I will use 150 mAH of CAPACITY of the battery. The current that you draw is 2A (for 28 sec.).> If the motor moves 13 times in a day then > > 13x 28 = 364 secs in a day.Wrong again. The motor will do 12 moves of 28 sec. and a big move of 3 minutes to go back, so it will move for 12* 28 sec + 180 sec = 516 sec. Notes about batteries: - read on the datasheet of the battery what is the rated current you can get. If you draw more current than what is rated you''l get less AH than what is rated. - If you want to keep the battery in good shape, keep at least 50% charge in it (so if battery is 4AH, don't use more than 2AH). - Temperature has a huge impact on battery capacity: the colder, the less capacity you can use. Bye Jack
Reply by ●December 23, 20152015-12-23
In article <c_ydnavWpaHBnefLnZ2dnUVZ5q2dnZ2d@giganews.com>, tim@seemywebsite.com says...> > On Tue, 22 Dec 2015 15:43:01 -0800, Jessica Shaw wrote: > > > Hi, > > > > I am using an accelerometer to calculate tilt. Its single axis control. > > Why 15 degrees an hour? > > Well, lessee. You're tracking the sun, and a day is 24 hours long, and > the sun scribes a circle in the sky once per day, and a day is 24 hours > long, and a circle has 360 degrees, and 360 / 24 = 15 and that ain't no > coincidence at all.Wont even go into depends on lattiude and time of year as well, as to work out longer and shorter arcs. But then again as the current draw has not been mapped out for the start/move/accelerate under different mechanical loadings and battery charge levels not much hope of sorting this. -- Paul Carpenter | paul@pcserviceselectronics.co.uk <http://www.pcserviceselectronics.co.uk/> PC Services <http://www.pcserviceselectronics.co.uk/pi/> Raspberry Pi Add-ons <http://www.pcserviceselectronics.co.uk/fonts/> Timing Diagram Font <http://www.badweb.org.uk/> For those web sites you hate
Reply by ●December 23, 20152015-12-23
On 12/23/2015 11:20 AM, Paul wrote:> In article <c_ydnavWpaHBnefLnZ2dnUVZ5q2dnZ2d@giganews.com>, > tim@seemywebsite.com says... >> >> On Tue, 22 Dec 2015 15:43:01 -0800, Jessica Shaw wrote: >> >>> Hi, >>> >>> I am using an accelerometer to calculate tilt. Its single axis control. >>> Why 15 degrees an hour? >> >> Well, lessee. You're tracking the sun, and a day is 24 hours long, and >> the sun scribes a circle in the sky once per day, and a day is 24 hours >> long, and a circle has 360 degrees, and 360 / 24 = 15 and that ain't no >> coincidence at all. > > Wont even go into depends on lattiude and time of year as well, as to > work out longer and shorter arcs. > > But then again as the current draw has not been mapped out for the > start/move/accelerate under different mechanical loadings and > battery charge levels not much hope of sorting this.No one said this is a sun tracker other than Tim who is likely mistaken. Or did I miss something? -- Rick
Reply by ●December 23, 20152015-12-23
On Wed, 23 Dec 2015 16:20:45 -0000, Paul <paul@pcserviceselectronics.co.uk> wrote:>In article <c_ydnavWpaHBnefLnZ2dnUVZ5q2dnZ2d@giganews.com>, >tim@seemywebsite.com says... >> >> On Tue, 22 Dec 2015 15:43:01 -0800, Jessica Shaw wrote: >> >> > Hi, >> > >> > I am using an accelerometer to calculate tilt. Its single axis control. >> > Why 15 degrees an hour? >> >> Well, lessee. You're tracking the sun, and a day is 24 hours long, and >> the sun scribes a circle in the sky once per day, and a day is 24 hours >> long, and a circle has 360 degrees, and 360 / 24 = 15 and that ain't no >> coincidence at all. > >Wont even go into depends on lattiude and time of year as well, as to >work out longer and shorter arcs.Just use equatorial mount, i.e. the panel axis is parallel to the axis of the earth. Adjust the panel axis according to your latitude. After that you just need one constant velocity mover to track the sun regardless of the day of year.>But then again as the current draw has not been mapped out for the >start/move/accelerate under different mechanical loadings and >battery charge levels not much hope of sorting this.The OP did not specify any feedback mechanism how to ensure accurate tracking. While I understand that a battery may be required to swing back the panel at night and perhaps during a few rainy days. In normal situations the panel would be quite sufficient to charge the battery. In the worst case after an extended rain period, turn the panel to the meridian (to the south in northern hemisphere) and when the sun appears, it will charge the battery sufficiently and the next day it will track the sun properly.
Reply by ●December 23, 20152015-12-23
I am using a single axis accelerometer to measure the tilt. Since, its single axis tracking. I am planning to only follow the elevation of the Sun not azimuth. So, I have Real Time clock on board and then I calculate the Sun's elevation for that particular day w.r.t time. So, the motor starts from 40 degrees ( mechanical assembly mounted to face the Sun at 40 degrees at Sun rise) and moves "x" ( elevation angle) degrees every hour until it reaches -40 degrees. I can calculate the Sun elevation w.r.t to time but how can I map the Sun's elevation values into my code to control the motor. I just turn ON and OFF the motor and monitor the tilt using accelerometer. I am not using PWM. I am just little confused about it
