Skip, you asked some really great questions in this post! I am not sure I
can answer them in other than general terms and thoughts. See the thoughts
interspersed with what you wrote below...
Subject: RE: Tires
> Hello Ed,
> Tire frontal area is a really big issue at any speed as drag increases
> greatly with increase in speed and the Cd is relatively high.......but how
> about considering this......the top of the tire is moving twice as fast as
> the car, relative to the wind, and the bottom of the tire is not moving at
> all relative to the wind (no slippage). So if you cover the top of the
> tire, most of the skin drag would be eliminated? Here's one to calculate,
> Mayf. The desired answer might be "How much of the tire to cover?" or what
> is the real drag of an open tire?
IMHO, the drag on a tires is complex to say the least. Maybe that is why
there is so little data available on the subject. The tire's drag
coefficient must be somwhat like sticking a flat plate into the wind and
those have high drag numbers. Add in rotational flow due to the boundary
layer on the tire and it is probably not solveable on your calculator. I had
a problem like this on a project once and it involved circular flow patterns
and circular shock waves. It was the inverse of this problem and it required
3D navier stokes equations and lots of computing power to solve. I know some
out there have raised an eyebrow about the top of the tire speed you
mentioned so let's make sure everyone is confident that this is true. First,
the center of the wheel/tire is moving at the same speed as the car. Then
you add the component due to rotation of the tire which is also the same as
the forward speed (if someone needs this off line, let me know and I'll give
your the math analysis that proves it). So the "relative wind" at the top of
a tire is 2 times the forward speed of the car. So how much "fender" to add?
This is a gut feel on my part but I would if the rules allowed it and there
was room in the fenderwell I would start at the 3:00 o'clock position cover
the top and front down to about the 8:00 o'clock position. This would leave
a about 150 degrees of the full wheel open. The back would be open to let
that pesky salt fly off of the tire and the front being closed would IMHO
help to reduce the circular air flow from mixing with the fenderwell flow.
This would cause a LOT of turbulence and turbulence is drag. A bigger
question is how close to the tire to make the fender? Well, remember that
the air attached to the tire is going like crazy and the air on the inner
side of the fender is stopped. Like the flow under your car. So there is a
high shear in this air and that also causes drag. So what's a mother to do?
Well, ever watched the world sports car classes? They have full fenders but
the fender tops have rear facing louvers over the top side of the tire. I
thinks this is to reduce the turbulent air inside the fender. The air flows
up and over the top of the fender and creates a low pressure at that point
and sucks out the air inside the fender leaving a low pressure low drag
condition inside the fender. Make sense? We are talking computational fluid
dynamics here and that button isn't on my calculator so I cannot solve...
But that is what I would do. Even on a full body car.
> The top of our tires were going 542 MPH at the "5". The top of Law's tires
> were going 700 MPH. Hammond's at 600. I wonder what sonic shock waves will
> do to tire tread...or.....how narrow a tire can we use and still be safe
> from excessive distortion? Seems we had a thread about flexing some months
> ago......wonder what happened? I seem to remember some sensitivity about
> it.....
Ok, one thing we again need to do is make sure everyone is on the same page
when we talk about supersonic conditions. The speed of sound is dependent
solely on the temperature of the air. Skeptics may write off line and I'll
send the math... But it breaks down to this:
a = 49.018 (sqrt of 459.6 + T). So on a day when the temperature is 95
degrees F then the speed of sound is 49.018 * sqrt (554.6) or a = 1154.4
ft/second. A side thought here...you'd think that it would be better to run
for the absolute LSR record (Thrust) when the temps were down because the
sonic velocity is down also. But remember that as temp goes down, the air
density goes up and hence drag goes up...it is a real trade off. Now that
we understand the definition of the speed of sound, MAch number is the
vehicle speed divided by the speed of sound. That's why when Thrust went
fast, they had to wait a second or two while the speed of sound for that run
was calculated so they could figure the Mach number... So how fast is 1154
ft/second? Well, it is 787 mph. So the toip of a tire on a 95 degree day
would have to be travelling 787 mph to get to Mach 1 and have a standing
normal shock wave on the top of the tire. But it is a very short shock wave
because it is only the air that is attached to the tire's boundary layer
that is going that fast. Just a few millimeters away from the tire, the air
is slowing because it is standing still either by virtue of the fender" or
free air stream. Now remember, this is all conjecture on my part! YMMV.
As to tires, well, I would be more concerned with the centrifugal force on
the tread section. Here is why: F = mass * tire radius * omega^2 (omega is 2
* pi * revs per sec of tire). As an example, lets say the tire is 28 inches
in diameter and weighs what? 24 pounds? Lets assume that the sidewalls of
the tire weigh 1/3 of the total weight, so the "tread section" then weighs
16 pounds. Lets pretend that this tire tread is 8 inches wide. At 28 inches
in diameter, the circumference of the tire is 2 * pi * 14 = 88 inches. So a
one inch wide strip of tread weighs about 0.18 pounds. Or, since our tread
is 8 inches wide, then one square inch of tread would weigh 0.023 pounds. So
now all we have to do is figure the rotational speed of the tire to figure
out how much centrifugal force there is on that 1 square inch of tread.
Let's assume 300 mph or 440 ft/sec. Divide this by the circumference of the
tire in feet and get 440 / 7.33 = 60 revolutions per second! Same as the
frequency of the AC power in your house. Pretty fast. Moving on, F =
0.023/32.2 * 1.16 ft * (2*pi* 60)^2
F = 0.023/32.2 * 1.16 * (376.99 * 376.99) = 117.75 pounds! on that little
one square inch! ( the 0.023 was the weight of the section, 32.2 converts it
to mass units, 1.16 is the tire radius in feet, pi = 3.1416, 60 was the
revolutions per second of the tire).
Since I used a one square inch of tread, this is the same as saying that the
tire is pressurized internally to nearly 120 psig when going 300 mph! And
that is on top of what you started with! So start with 60 psig then you have
the equivalent of 180 psig acting on your tires! The tread is under a
tremendous strain. Because most tires are radial (are the Goodyears? I think
so), the sidewall cord keepd the edge of the tread from expanding as much as
the center of the tread so the tire takes on an oval shape. Reducing the
contact patch and reducing rolling resistance a whole hell of a lot! IMHO.
And if you are adventurous, you can make a stab at how the salt stays in a
treaded tire at these speeds...I don't think it does, I think it comes off
nearly as soon as it sticks on the tire...
Was this mental masturbation enough for everyone? Fun for me, was it good
for you?
mayf, the red necked ignorant desert rat in Pahrump where the new house
power is on, water runs from the well, counter tops, sinks, and associated
hardware is in, base boards are going in, the shop under sink hot water
heater is in (main house has tankless water heaters), toilets are all in and
have water in them, getting closer....Maybe by the end of next week?
nah...bet it is 2 more weeks...
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