An Elementary Treatise on Quaternions |
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Page 141
... envelop of the planes of inter- section . Let ( p - a ) 2 = a2 be the first sphere , i.e. p2 - 2Sap = 0 . One of the others is p2-28ap = 0 , where The plane of intersection is Hence , for the envelop , ( see next Chapter , ) Sapa = 1 ...
... envelop of the planes of inter- section . Let ( p - a ) 2 = a2 be the first sphere , i.e. p2 - 2Sap = 0 . One of the others is p2-28ap = 0 , where The plane of intersection is Hence , for the envelop , ( see next Chapter , ) Sapa = 1 ...
Page 152
... envelop of all non - central plane sections of an ellip- soid whose area is constant . 8. Find the locus of the ... envelop of the planes of contact of tangent planes drawn to an ellipsoid from points of a concentric sphere . Find the ...
... envelop of all non - central plane sections of an ellip- soid whose area is constant . 8. Find the locus of the ... envelop of the planes of contact of tangent planes drawn to an ellipsoid from points of a concentric sphere . Find the ...
Page 164
... = p + up = & + up ' , 300. ] This leads us to the consideration of envelops generally , and the process just employed may easily be extended to the problem of finding the envelop of a series of surfaces whose 164 [ 298 . QUATERNIONS .
... = p + up = & + up ' , 300. ] This leads us to the consideration of envelops generally , and the process just employed may easily be extended to the problem of finding the envelop of a series of surfaces whose 164 [ 298 . QUATERNIONS .
Page 166
... envelops . 303. ] Find the envelop of a plane whose distance from the origin is constant . Here with the condition Hence , by last section , and therefore or Sap - c , Ta = 1 . Γρα Vpa = 0 , p = ca , Τρ Tp = c , the sphere of radius c ...
... envelops . 303. ] Find the envelop of a plane whose distance from the origin is constant . Here with the condition Hence , by last section , and therefore or Sap - c , Ta = 1 . Γρα Vpa = 0 , p = ca , Τρ Tp = c , the sphere of radius c ...
Page 167
... envelop if ß and be constant , and a be subject to the one scalar condition = 1 . Ta The process of § 302 gives , by inspection , pSap + VBp = xa . Operating by . S.a , we get S2ap + S.aßp — x , which gives S.aßp = x + b . But , by ...
... envelop if ß and be constant , and a be subject to the one scalar condition = 1 . Ta The process of § 302 gives , by inspection , pSap + VBp = xa . Operating by . S.a , we get S2ap + S.aßp — x , which gives S.aßp = x + b . But , by ...
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a₁ axes axis Cartesian centre of inertia Chapter circle cloth cone conjugate constant cöordinates coplanar curvature curve developable surface diameters differential direction drawn easily ellipsoid envelop equal evidently expression Extra fcap extremity fcap Find the equation Find the locus given equation given line given vectors gives Hamilton Hence hodograph integral intersection last section length linear and vector locus normal obviously once operator origin osculating osculating plane P₁ parabola parallel perpendicular properties quaternion radius rectangular represents result right angles rotation S.aßy Saß scalar scalar equations second order self-conjugate shew solution sphere spherical straight line suppose surface surface of revolution tangent plane Taylor's Theorem tensor theorem three vectors triangle unit-vectors Vaß vector function versor w₁ write written Τρ φρ