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Question-47824




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.
$${Find}\:{the}\:{equation}\:{of}\:{the}\:{shadow} \\ $$$${curve}\:{of}\:{a}\:{sphere}\:{of}\:{radius}\:{R} \\ $$$${mounted}\:{on}\:{a}\:{pole}\:{of}\:{height}\:{h}. \\ $$
Commented by ajfour last updated on 15/Nov/18
 i think BM > DM.
$$\:{i}\:{think}\:{BM}\:>\:{DM}. \\ $$
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  _________________________.
$${S}\left(\mathrm{0},−{d},{H}\right)\:\:;\:{let}\:{E}\:{be}\:{center}\:{of} \\ $$$${sphere}:\:\:\:{E}\left(\mathrm{0},\mathrm{0},{h}+{R}\right)\equiv\left(\mathrm{0},\mathrm{0},{z}_{\mathrm{0}} \right) \\ $$$${let}\:{a}\:{ray}\:{grazing}\:{the}\:{sphere}\:{hit} \\ $$$${the}\:{shadow}-{curve}\:{boundary}\:{at} \\ $$$$\left({p},{q}\right). \\ $$$${Eq}.\:{of}\:{such}\:{a}\:{ray}: \\ $$$$\bar {{r}}=−{d}\hat {{j}}+{H}\hat {{k}}+\lambda\left({p}\hat {{i}}+\left({q}+{d}\right)\hat {{j}}−{H}\hat {{k}}\right) \\ $$$$\bot\:{distance}\:{of}\:\bar {{v}}\:{from}\:{a}\:{line} \\ $$$$\:\:\bar {{r}}\:=\bar {{a}}+\lambda\bar {{b}}\:,\:{we}\:{need}\:{to}\:{obtain}\:{this} \\ $$$${first}. \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${let}\:{foot}\:{of}\:\bot\:{from}\:\bar {{v}}\:{to}\:{line}\:{be} \\ $$$$\:\:\bar {{r}}\:=\:\bar {{a}}+\lambda_{\mathrm{0}} \bar {{b}} \\ $$$$\Rightarrow\:\:\left(\bar {{a}}+\lambda_{\mathrm{0}} \bar {{b}}−\bar {{v}}\right).\bar {{b}}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\lambda_{\mathrm{0}} =\frac{\left(\bar {{v}}−\bar {{a}}\right).\bar {{b}}}{\bar {{b}}.\bar {{b}}} \\ $$$${hence}\:\bot\:{distance}\:{is}\: \\ $$$$\:\:\:\:\:{d}\:=\:\mid\bar {{a}}+\lambda_{\mathrm{0}} \bar {{b}}−\bar {{v}}\mid \\ $$$$\:\:\:\:\:\:\:\:=\:\mid\bar {{a}}−\bar {{v}}+\left(\frac{\left(\bar {{v}}−\bar {{a}}\right).\bar {{b}}}{\bar {{b}}.\bar {{b}}}\right)\bar {{b}}\mid \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ \\ $$$${Now}\:\bot\:{distance}\:{of}\:{ray}\:{from} \\ $$$${center}\:{of}\:{sphere}\:{E}\:{is}\:{R}.\:{So} \\ $$$$\:\mid−{d}\hat {{j}}+{H}\hat {{k}}+\frac{\left[{d}\hat {{j}}+\left({h}+{R}−{H}\right)\hat {{k}}\right].\left[{p}\hat {{i}}+\left({q}+{d}\right)\hat {{j}}−{H}\hat {{k}}\right]\left({p}\hat {{i}}+\left({q}+{d}\right)\hat {{j}}−{H}\hat {{k}}\right]}{{p}^{\mathrm{2}} +\left({q}+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }−\left({h}+{R}\right)\hat {{k}}\mid={R} \\ $$$$\mid−{d}\hat {{j}}+{H}\hat {{k}}+\frac{\left[{d}\left({y}+{d}\right)−{H}\left({h}+{R}−{H}\right)\right]\left({x}\hat {{i}}+\left({y}+{d}\right)\hat {{j}}−{H}\hat {{k}}\right]}{{x}^{\mathrm{2}} +\left({y}+{d}\right)^{\mathrm{2}} +{H}^{\mathrm{2}} }−\left({h}+{R}\right)\hat {{k}}\mid={R} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. \\ $$
Commented by ajfour last updated on 15/Nov/18
MrW Sir, any polar method to  solve this question ?
$${MrW}\:{Sir},\:{any}\:{polar}\:{method}\:{to} \\ $$$${solve}\:{this}\:{question}\:? \\ $$
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.
$$\mathrm{Sir},\:\mathrm{Do}\:\mathrm{you}\:\mathrm{use}\:\mathrm{Lekh}\:\mathrm{diagram}\:\mathrm{to}\:\mathrm{draw}\:\mathrm{this}\:\mathrm{or}\:\mathrm{another}\:\mathrm{app}.\: \\ $$
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.
$${Due}\:{to}\:{the}\:{problem}\:{with}\:{my}\:{new}\: \\ $$$${smartphone}\:{I}\:{won}'{t}\:{be}\:{able}\:{to}\:{type} \\ $$$${the}\:{complete}\:{solution}.\:{I}\:{just}\:{can}\:{not} \\ $$$${use}\:{the}\:{app}\:{to}\:{edit}\:{formulas}\:{because} \\ $$$${the}\:{last}\:{row}\:{of}\:{the}\:{menu}\:{is}\:{not}\:{available}. \\ $$$${I}\:{hope}\:{TinkuTara}\:{could}\:{help}\:{me}\:{soon} \\ $$$${to}\:{solve}\:{this}\:{problem}. \\ $$$$ \\ $$$${My}\:{solution}\:{is}\:{quite}\:{easy}.\:{It}\:{uses}\:{that} \\ $$$${what}\:{we}\:{already}\:{know}\:{from}\:\:{Q}\mathrm{44017}. \\ $$$$ \\ $$$${The}\:{shadow}\:{of}\:{the}\:{sphere}\:{on}\:{the}\:{ground} \\ $$$${is}\:{an}\:{oblique}\:{section}\:{of}\:{the}\:{right}\:{cone} \\ $$$${S}−{BPQ}.\:{From}\:{Q}\mathrm{44017}\:{we}\:{know}\:{that}\:{it}'{s} \\ $$$${an}\:{ellipse}.\:{The}\:{parameters}\:{a},{b}\:{for}\:{this} \\ $$$${ellipse}\:{can}\:{be}\:{calculated}\:{with}\:{the} \\ $$$${formulas}\:{for}\:{a}\:{and}\:{b}\:{in}\:{Q}\mathrm{44017}.\:{When} \\ $$$${we}\:{have}\:{them},\:{the}\:{rest}\:{is}\:{clear}. \\ $$
Commented by ajfour last updated on 16/Nov/18
Thanks for the diagrams, Sir.  I′ll try to obtain the equation  this way.
$${Thanks}\:{for}\:{the}\:{diagrams},\:{Sir}. \\ $$$${I}'{ll}\:{try}\:{to}\:{obtain}\:{the}\:{equation} \\ $$$${this}\:{way}. \\ $$
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).
$$\:\:\:\:\underset{−} {{Kindly}\:{allow}\:{a}\:{doubt}} \\ $$$${I}\:{doubt}\:{if}\:{its}\:{an}\:{ellipse}; \\ $$$${corresponding}\:{to}\:{center}\:{of} \\ $$$${object}\:{center},\:{M}\:{is}\:{the}\:{shadow} \\ $$$${center};\:{and}\:{since}\:{projection}\:{of} \\ $$$${QP}\:{on}\:{ground}\:{isDM}\:\left({less}\right. \\ $$$$\left.{magnified}\right)\:\:,\:{while}\:{projection} \\ $$$${of}\:{BP}\:{becomes}\:{BM}\left({magnified}\right. \\ $$$$\left.{more}\right)\:,\:{so}\:{the}\:{two}\:{semi}-{major} \\ $$$${axis}\:{parts},\:\:{DM}\:{and}\:{MB}\:{get} \\ $$$${unequal},\:{M}\:{still}\:{being}\:{the}\:{center} \\ $$$${of}\:{shadow}\:\left({not}\:{a}\:{proper}\:{ellipse}\right). \\ $$
Commented by ajfour last updated on 16/Nov/18
I use Lekh Diagram only.
$${I}\:{use}\:{Lekh}\:{Diagram}\:{only}. \\ $$
Commented by Tawa1 last updated on 16/Nov/18
Thanks sir.  I mean sir mrW diagram
$$\mathrm{Thanks}\:\mathrm{sir}.\:\:\mathrm{I}\:\mathrm{mean}\:\mathrm{sir}\:\mathrm{mrW}\:\mathrm{diagram} \\ $$
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.
$${ajfour}\:{sir}: \\ $$$${M}\:{is}\:{the}\:{shadow}\:{of}\:{the}\:{center}\:{of}\:{the} \\ $$$${sphere},\:{but}\:{it}\:{is}\:{not}\:{the}\:{center}\:{of}\:{the} \\ $$$${shadow}.\:{DM}<{MB}\:{is}\:{correct}. \\ $$$${the}\:{center}\:{of}\:{the}\:{shadow}\:{is}\:{point}\:{N}, \\ $$$${it}\:{is}\:{there}\:{where}\:{the}\:{shadow}\:{widest}. \\ $$$${since}\:{the}\:{shadow}\:{is}\:{an}\:{oblique}\:{section} \\ $$$${of}\:{a}\:{cone},\:{so}\:{it}\:{is}\:{an}\:{ellipse}.\:{this}\:{is} \\ $$$${sure}. \\ $$
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.
$${to}\:{tawa}\mathrm{1}\:{sir}: \\ $$$${I}\:{don}'{t}\:{use}\:{a}\:{special}\:{app}\:{to}\:{draw}\:{my} \\ $$$${diagrams}.\:{Sometimes}\:{I}\:{draw}\:{them} \\ $$$${as}\:{comments}\:{in}\:{a}\:{pdf}\:{document}\:{and} \\ $$$${make}\:{screen}\:{captures}\:{from}\:{them}. \\ $$
Commented by mr W last updated on 16/Nov/18
Commented by Tawa1 last updated on 16/Nov/18
Alright sir. Thanks.
$$\mathrm{Alright}\:\mathrm{sir}.\:\mathrm{Thanks}.\: \\ $$
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!
$${You}\:{are}\:{great}\:{Sir},\:{exactly}\:{what} \\ $$$${i}\:{must}\:{have}\:{first}! \\ $$

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