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Question Number 47824 by ajfour last updated on 15/Nov/18

Commented by ajfour last updated on 15/Nov/18

Find the equation of the shadow curve of a sphere of radius R mounted on a pole of height h.

FindtheequationoftheshadowcurveofasphereofradiusRmountedonapoleofheighth.

Commented by ajfour last updated on 15/Nov/18

i think BM > DM.

ithinkBM>DM.

Answered by ajfour last updated on 16/Nov/18

S(0,−d,H) ; let E be center of sphere: E(0,0,h+R)≡(0,0,z_0 ) let a ray grazing the sphere hit the shadow-curve boundary at (p,q). Eq. of such a ray: r^� =−dj^� +Hk^� +λ(pi^� +(q+d)j^� −Hk^� ) ⊥ distance of v^� from a line r^� =a^� +λb^� , we need to obtain this first. _________________________ let foot of ⊥ from v^� to line be r^� = a^� +λ_0 b^� ⇒ (a^� +λ_0 b^� −v^� ).b^� = 0 ⇒ λ_0 =(((v^� −a^� ).b^� )/(b^� .b^� )) hence ⊥ distance is d = ∣a^� +λ_0 b^� −v^� ∣ = ∣a^� −v^� +((((v^� −a^� ).b^� )/(b^� .b^� )))b^� ∣ ________________________ Now ⊥ distance of ray from center of sphere E is R. So ∣−dj^� +Hk^� +(([dj^� +(h+R−H)k^� ].[pi^� +(q+d)j^� −Hk^� ](pi^� +(q+d)j^� −Hk^� ])/(p^2 +(q+d)^2 +H^2 ))−(h+R)k^� ∣=R ∣−dj^� +Hk^� +(([d(y+d)−H(h+R−H)](xi^� +(y+d)j^� −Hk^� ])/(x^2 +(y+d)^2 +H^2 ))−(h+R)k^� ∣=R _________________________.

S(0,−d,H);letEbecenterofsphere:E(0,0,h+R)≡(0,0,z0)letaraygrazingthespherehittheshadow−curveboundaryat(p,q).Eq.ofsucharay:r¯=−dj^+Hk^+λ(pi^+(q+d)j^−Hk^)⊥distanceofv¯fromaliner¯=a¯+λb¯,weneedtoobtainthisfirst._________________________letfootof⊥fromv¯tolineber¯=a¯+λ0b¯⇒(a¯+λ0b¯−v¯).b¯=0⇒λ0=(v¯−a¯).b¯b¯.b¯hence⊥distanceisd=∣a¯+λ0b¯−v¯∣=∣a¯−v¯+((v¯−a¯).b¯b¯.b¯)b¯∣________________________Now⊥distanceofrayfromcenterofsphereEisR.So∣−dj^+Hk^+[dj^+(h+R−H)k^].[pi^+(q+d)j^−Hk^](pi^+(q+d)j^−Hk^]p2+(q+d)2+H2−(h+R)k^∣=R∣−dj^+Hk^+[d(y+d)−H(h+R−H)](xi^+(y+d)j^−Hk^]x2+(y+d)2+H2−(h+R)k^∣=R_________________________.

Commented by ajfour last updated on 15/Nov/18

MrW Sir, any polar method to solve this question ?

MrWSir,anypolarmethodtosolvethisquestion?

Answered by mr W last updated on 15/Nov/18

Commented by Tawa1 last updated on 16/Nov/18

Sir, Do you use Lekh diagram to draw this or another app.

Sir,DoyouuseLekhdiagramtodrawthisoranotherapp.

Commented by mr W last updated on 16/Nov/18

Due to the problem with my new smartphone I won′t be able to type the complete solution. I just can not use the app to edit formulas because the last row of the menu is not available. I hope TinkuTara could help me soon to solve this problem. My solution is quite easy. It uses that what we already know from Q44017. The shadow of the sphere on the ground is an oblique section of the right cone S−BPQ. From Q44017 we know that it′s an ellipse. The parameters a,b for this ellipse can be calculated with the formulas for a and b in Q44017. When we have them, the rest is clear.

DuetotheproblemwithmynewsmartphoneIwon′tbeabletotypethecompletesolution.Ijustcannotusetheapptoeditformulasbecausethelastrowofthemenuisnotavailable.IhopeTinkuTaracouldhelpmesoontosolvethisproblem.Mysolutionisquiteeasy.ItusesthatwhatwealreadyknowfromQ44017.TheshadowofthesphereonthegroundisanobliquesectionoftherightconeS−BPQ.FromQ44017weknowthatit′sanellipse.Theparametersa,bforthisellipsecanbecalculatedwiththeformulasforaandbinQ44017.Whenwehavethem,therestisclear.

Commented by ajfour last updated on 16/Nov/18

Thanks for the diagrams, Sir. I′ll try to obtain the equation this way.

Thanksforthediagrams,Sir.I′lltrytoobtaintheequationthisway.

Commented by ajfour last updated on 16/Nov/18

Kindly allow a doubt_(−) I doubt if its an ellipse; corresponding to center of object center, M is the shadow center; and since projection of QP on ground isDM (less magnified) , while projection of BP becomes BM(magnified more) , so the two semi-major axis parts, DM and MB get unequal, M still being the center of shadow (not a proper ellipse).

Kindlyallowadoubt−Idoubtifitsanellipse;correspondingtocenterofobjectcenter,Mistheshadowcenter;andsinceprojectionofQPongroundisDM(lessmagnified),whileprojectionofBPbecomesBM(magnifiedmore),sothetwosemi−majoraxisparts,DMandMBgetunequal,Mstillbeingthecenterofshadow(notaproperellipse).

Commented by ajfour last updated on 16/Nov/18

I use Lekh Diagram only.

IuseLekhDiagramonly.

Commented by Tawa1 last updated on 16/Nov/18

Thanks sir. I mean sir mrW diagram

Thankssir.ImeansirmrWdiagram

Commented by mr W last updated on 16/Nov/18

ajfour sir: M is the shadow of the center of the sphere, but it is not the center of the shadow. DM<MB is correct. the center of the shadow is point N, it is there where the shadow widest. since the shadow is an oblique section of a cone, so it is an ellipse. this is sure.

ajfoursir:Mistheshadowofthecenterofthesphere,butitisnotthecenteroftheshadow.DM<MBiscorrect.thecenteroftheshadowispointN,itistherewheretheshadowwidest.sincetheshadowisanobliquesectionofacone,soitisanellipse.thisissure.

Commented by mr W last updated on 16/Nov/18

to tawa1 sir: I don′t use a special app to draw my diagrams. Sometimes I draw them as comments in a pdf document and make screen captures from them.

totawa1sir:Idon′tuseaspecialapptodrawmydiagrams.SometimesIdrawthemascommentsinapdfdocumentandmakescreencapturesfromthem.

Commented by mr W last updated on 16/Nov/18

Commented by Tawa1 last updated on 16/Nov/18

Alright sir. Thanks.

Alrightsir.Thanks.

Commented by mr W last updated on 16/Nov/18

Commented by ajfour last updated on 16/Nov/18

You are great Sir, exactly what i must have first!

YouaregreatSir,exactlywhatimusthavefirst!

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