ANGLE OF ORIGIN OF VESSELS 293 



right angles (say at D). For the purposes of this calculation 

 let us consider that the least loss of power occurs when the branch 

 originates at X which is x units from Z), making an angle of 6 

 with the main trunk. Then the distance from X to C would be 

 \/# 2 -f/i 2 (hypotenuse of right-angled triangle). 



Assuming that loss of pressure is due to friction on the walls 

 of the vessels, then it will be directly proportional to their lengths 

 and indirectly proportional to their radii (e.g. main trunk =R, 

 branch =r) ; 



XC AX 



i.e. loss is proportional to - +-^- 



T K 



If the whole distance from A to D be put =6, then AX=bx. 



c , ... ,. v# 2 +ft 2 b-x 



. Substituting, we have - + , 



T K 



multiplying by Rr gives us the value 



-x)r. 



Differentiating and equating to zero we obtain a value for x 

 which makes S a minimum. 



Thus 



d.x 



XD 



That is, the angle of origin required is such that its cosine is 

 numerically equal to the radius of the branch divided by the radius 

 of the main trunk. 



The size of the angle of origin is governed neither by the radius 

 of the branch vessel nor by the radius of the main vessel but by 

 the ratio of these two quantities. For any particular value of the 

 ratio r/R, we have therefore a constant value of 6 ; that is, all 

 branches of equal radius will be equally inclined to the main 

 artery. 



(1) In particular, if the artery bifurcates into two equal 

 branches, the angles of bifurcation will be equal. 



(2) If r is so small compared with R that the amount of blood 



going to the branch is almost negligible, then cos 6 tends to 

 be infinitely small, i.e. angle will be close to 90. 



