What would happen to LP ( Polar) if one has to find the insulation of a locality in the northern hemisphere with declination delta having a negative angle value ? Will LP (polar) be 0 or or sin(negative delta) ?
At 24:22 : min I couldn't understand how '''' Lz =Lm*cos(fi )+ Lp*sin(fi)''' ,... what I know is Lm=Lz*cos(fi) & Lp=Lz*sin(fi)'''''o'''' and Lz=Lm+Lp (vectro sum),o correct me if I where wrong please
The insolation vector is defined as L = L_S*Shat + L_E*Ehat + L_Z*Zhat = L_m*mhat + L_e*ehat + L_p*phat. Here, "hat" terms are unit vectors. The unit vectors in earth-centric frame can be expressed in terms of local-centric frame as mhat = Zhat*cos(phi) + Shat*sin(phi), phat = Zhat*sin(phi) - Shat*cos(phi), and ehat = Ehat. Substituting these values into insolation vector equation we get: L = L_S*Shat + L_E*Ehat + L_Z*Zhat = L_m*(Zhat*cos(phi) + Shat*sin(phi)) + L_e*Ehat + L_p*(Zhat*sin(phi) - Shat*cos(phi)). Collecting Zhat terms on the RHS, we can solve for L_Z = L_m*cos(phi) + L_p*sin(phi). Hope that helps.
@@victoryu2025 Thanks Victor. Is there a video or textbox that maybe explains what you described for us visual learners? I also struggled with how the derivation at 24:22 was arrived at.
@@desousaj99 Are you confused for the same reason as the original comment from Bilal Bilal? It seems like a few other comments also expressed doubt about the derivation at 24:22. I will make a clarification video and post the link here shortly.
Yes Victor. I understand what Bilal Bilal is noting above. That's how I would have thought to resolve the Lz into Lm and Lp components. Looking forward to your video. Have been trying to sort it out the last couple of days and I am stumped. Thanks in advance.
excellent explanation, thanks !
What would happen to LP ( Polar) if one has to find the insulation of a locality in the northern hemisphere with declination delta having a negative angle value ? Will LP (polar) be 0 or or sin(negative delta) ?
please suggest test book where this concept is available
I reckon there is a small mathematical error in derivation. At 24:22 How Lz = Lm cos pi + Lp sin pi? Any supporting text book for this derivation?
Here is a video I made explaining this derivation in more detail: ua-cam.com/video/11kpnpxqO5s/v-deo.html
very good explanation sir you deserve it
Punk
thank you sir for all what you have explained
At 24:22 : min I couldn't understand how '''' Lz =Lm*cos(fi )+ Lp*sin(fi)''' ,... what I know is Lm=Lz*cos(fi) & Lp=Lz*sin(fi)'''''o''''
and Lz=Lm+Lp (vectro sum),o
correct me if I where wrong please
The insolation vector is defined as L = L_S*Shat + L_E*Ehat + L_Z*Zhat = L_m*mhat + L_e*ehat + L_p*phat. Here, "hat" terms are unit vectors. The unit vectors in earth-centric frame can be expressed in terms of local-centric frame as mhat = Zhat*cos(phi) + Shat*sin(phi), phat = Zhat*sin(phi) - Shat*cos(phi), and ehat = Ehat. Substituting these values into insolation vector equation we get: L = L_S*Shat + L_E*Ehat + L_Z*Zhat = L_m*(Zhat*cos(phi) + Shat*sin(phi)) + L_e*Ehat + L_p*(Zhat*sin(phi) - Shat*cos(phi)). Collecting Zhat terms on the RHS, we can solve for L_Z = L_m*cos(phi) + L_p*sin(phi). Hope that helps.
@@victoryu2025 Thanks Victor. Is there a video or textbox that maybe explains what you described for us visual learners? I also struggled with how the derivation at 24:22 was arrived at.
@@desousaj99 Are you confused for the same reason as the original comment from Bilal Bilal? It seems like a few other comments also expressed doubt about the derivation at 24:22. I will make a clarification video and post the link here shortly.
Yes Victor. I understand what Bilal Bilal is noting above. That's how I would have thought to resolve the Lz into Lm and Lp components. Looking forward to your video. Have been trying to sort it out the last couple of days and I am stumped. Thanks in advance.
@@desousaj99 Here is a link to the video: ua-cam.com/video/11kpnpxqO5s/v-deo.html
great explanation sir