Spokes

## Arthur's Spokes Issue

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Numbers:

#### Front Wheel Spokes

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r = 1 i n {\displaystyle r=1in}
h = 9.5 i n {\displaystyle h=9.5in}
V = π ∗ r 2 ∗ h {\displaystyle V=\pi *r^{2}*h}
Volume of each spoke found to be  29.83 i n 3 {\displaystyle 29.83in^{3}} , assuming pi to 2 decimal places.
Assuming the following:
T = 20 o C {\displaystyle T=20^{o}C}
P = 101.3 k P a {\displaystyle P=101.3kPa}
ρ = 1.2041 k g / m 3 {\displaystyle \rho =1.2041kg/m^{3}}
Convert the volume to metric units:
V = 29.83 i n 3 ∗ ( 2.54 c m i n ) 3 {\displaystyle V=29.83in^{3}*({\frac {2.54cm}{in}})^{3}}
V = 483.08 c m 3 {\displaystyle V=483.08cm^{3}}
V = 483.08 c m 3 ∗ ( m 100 c m ) 3 {\displaystyle V=483.08cm^{3}*({\frac {m}{100cm}})^{3}}
V = 4.84 X 10 − 4 m 3 {\displaystyle V=4.84X10^{-4}m^{3}}
Using the stated assumptions, the estimated weight due to spokes having no air:
m = ρ ∗ V {\displaystyle m=\rho *V}
m = ( 1.2041 k g / m 3 ) ∗ ( 4.84 X 10 − 4 m 3 ) {\displaystyle m=(1.2041kg/m^{3})*(4.84X10^{-4}m^{3})}
m = 5.81 X 10 − 4 k g {\displaystyle m=5.81X10^{-4}kg}
m p e r w h e e l = 5 ∗ m {\displaystyle m_{perwheel}=5*m}
m p e r w h e e l = 2.91 X 10 − 3 k g = 2.91 g {\displaystyle m_{perwheel}=2.91X10^{-3}kg=2.91g}
Assuming the rims weigh 20 kg, this decrease in weight can be disregarded for these purposes. Are we 100% sure these are not already a vacuum?

Look these over, let me know. If you are 100% sure on the above question, I will continue calculations for you.

Ms. C

AJ: I am 100% certain that there is not already a vacuum. There is air. I realize that 2.91 g is a negligible amount of weight. I am trying to figure out the forces exerted on the wheel at high speed.

Ms.C: Are the spokes solid? Most in industry are.... we can communicate through here as well....

AJ: I am completely certain that they are not solid all the way through. They are hollow, and the interior dimensions of the rims are conservative, I believe there is more mass inside each rim. I have seen cut outs of the rim, I have seen the manufacturing process. The rims are advertised as hollow, they are trademarked as aircore rims. I have been trying to find information on the effects of multiplied gravitational forces on air. I have had no luck. I suspect that there is a exponential effect. As the wheel speeds up, the centrifugal force increases, as the force increases, the air pressure increases, which increases the mass of air that the centrifugal force acts upon.

Your thoughts?

Here is a link to a photo of a cutaway diagram for a similar rim. http://veight.de/wp-content/uploads/2012/03/2013_Koenigsegg_Agera_R_Carbon_Fiber_Wheel_Hollow.jpg This is an incomplete diagram it, although it is a center locking wheel it fails to show hub pegs holes at the center of the rim. While hub pegs and the rim have a very tight fit they are not air tight.

On a side note dialogue with the hiring manager at Koenigsegg is going well.