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 one year ago
RCF=[(rpmx2pi)/60]^2 x r/980 How would I rearrange this to solve for rpm? Please show steps! Thanks :
 one year ago
RCF=[(rpmx2pi)/60]^2 x r/980 How would I rearrange this to solve for rpm? Please show steps! Thanks :

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aaronq
 one year ago
Best ResponseYou've already chosen the best response.1RCF=[(rpmx2pi)/60]^2 x r/980 (980)(RCF)/r=[(rpmx2pi)/60]^2 ((980)(RCF)/r)^(1/2)=[(rpmx2pi)/60] [((980)(RCF)/r)^(1/2)](60)=(rpmx2pi) [((980)(RCF)/r)^(1/2)](60/2pi)=rpm

clarkexo
 one year ago
Best ResponseYou've already chosen the best response.0using the nomograph I'm not getting the same answer?

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1what are the values for r and RCF

clarkexo
 one year ago
Best ResponseYou've already chosen the best response.07 cm for the radius and 600 RCF

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1i got rpm = 2767.64 are you sure everything is in the same units?

clarkexo
 one year ago
Best ResponseYou've already chosen the best response.0Can u please go over the steps of how your calculating it? Perhaps i'm doing it wrong..the answer using the graph is aprox 2760

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1RCF=[(rpmx2pi)/60]^2 x r/980 multiply both sides by reciprocal of r/980...which is 980/r (980)(RCF)/r=[(rpmx2pi)/60]^2 take the square root of both sides ((980)(RCF)/r)^(1/2)=[(rpmx2pi)/60] multiply both sides by reciprocal of 2pi/60... which is 60/2pi [((980)(RCF)/r)^(1/2)](60/2pi)=rpm RCF=600, r=7 [(980)(600)/7]^(1/2)[60/2pi]=rpm multiply: 980x600 [(588000/7]^(1/2)[60/2pi]=rpm divide: 588000/7 [(84000]^(1/2)[60/2pi]=rpm Take the square root of 84000 [289.827][60/2pi]=rpm multply 289.827 by 60 17389.65/2pi=rpm divide 17389.65 by 2 8694.826/pi=rpm divide 8694.826 by pi 2767.64909=rpm

clarkexo
 one year ago
Best ResponseYou've already chosen the best response.0Thank you so much!! I forgot to square root! You're amazing! Thanks again :)

aaronq
 one year ago
Best ResponseYou've already chosen the best response.1no prob! glad to be of help
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