how to convert radiation on horizontal plane to an inclined one

[sensij]

I think this is the one you are looking for.
The formula is TRUE for any incident as far as you have the incident sun angle.

You are going to need to have the hourly sun angle "alpha" to do the calculation.

Those equations provide trig relationships that are only valid at solar noon each day, and assume all irradiation is direct normal, in other words, directly from the disc of the sun. They do very little to help with the more general solution that the OP appeared to want, using data that a typical weather station would provide.

[myfriendSUN]

Those equations provide trig relationships that are only valid at solar noon each day, and assume all irradiation is direct normal, in other words, directly from the disc of the sun. They do very little to help with the more general solution that the OP appeared to want, using data that a typical weather station would provide.

For the OP to get what he want he needs to have the sun angle at each hour of the day and plug it in the formula.
The calculator from the link only works at solar noon each day.
The formula is true ALL the time.

[myfriendSUN]

sun angle alpha (at solar noon) = 90 - latitude + declination

declination depends on date of the year

Can we measure the sun angle at any incident or at each hour ???

http://pveducation.org/pvcdrom/properties-of-sunlight/sun-position-high-accuracy

May be use the sun's position calculator to find the sun angle at any incident.

[J.P.M.]

My friendSUN:

No offense, but you're out of your league on this one and demonstrating that a little knowledge can be a dangerous thing.

I'd respectfully suggest you check out some of the references provided in this post for solar position, incident angle information and info on the various components of solar radiation that pass through the earth's atmosphere. Also, you seem to have gotten off track that the goal is solar radiation on a surface, as well as that idea that the total POA rad. is more than simply the beam component.

[myfriendSUN]

My friendSUN:

No offense, but you're out of your league on this one and demonstrating that a little knowledge can be a dangerous thing.

I'd respectfully suggest you check out some of the references provided in this post for solar position, incident angle information and info on the various components of solar radiation that pass through the earth's atmosphere. Also, you seem to have gotten off track that the goal is solar radiation on a surface, as well as that idea that the total POA rad. is more than simply the beam component.

Mr. J.P.M.

I an a newbie to solar here, just try to learn something here.
I admitted I might be out of league on this one.
Just feel free to tell me which phase I said is wrong and I would be happy to dig out more info.

I just try to answer the OP's question directly for what he wants to do.
I don't care and don't know whether he is supposed to do what he wants to do.
I know there are a lot of other factors to consider, not just the beam component.
But that's what the OP asked for:
I have data for the hourly radiation on a horizontal plane and weather conditions for a location for a few years. I need to convert the radiation on the horizontal plane to the radiation on an inclined plane.
And the answer to his question would be the formula I posted.

[J.P.M.]

Mr. J.P.M.

I an a newbie to solar here, just try to learn something here.
I admitted I might be out of league on this one.
Just feel free to tell me which phase I said is wrong and I would be happy to dig out more info.

I just try to answer the OP's question directly for what he wants to do.
I don't care and don't know whether he is supposed to do what he wants to do.
I know there are a lot of other factors to consider, not just the beam component.
But that's what the OP asked for:
I have data for the hourly radiation on a horizontal plane and weather conditions for a location for a few years. I need to convert the radiation on the horizontal plane to the radiation on an inclined plane.
And the answer to his question would be the formula I posted.

In spite of what you may think, the answer to the OP's question is not to be found in the formula you posted.

This subject has been addressed in a very thorough manner in other posts to this forum.

Read them, study them, and as a newbie, learn. I can point you in the direction of self help. I'm a poor teacher and a worse spoon feeder. Both waste my and most everyone else's time and good will.