Saturday, November 21, 2009

Round tube vs streamline tube

I'll preface this by saying I'm a big fan of "order of magnitude" approximations... i.e. spending an hour to get an answer that's 85-90% correct rather than spending a week to get an answer that's 95% correct. For any non-engineers out there.. there's really no such thing as an exact answer in a lot of this stuff. Just a question of how accurate you want to get. In my opinion some people get too caught up too early with trying to find an exact solution.

This goes along with my long-standing belief that 'perfect' is the enemy of 'good.'

It's useful to do this stuff to get an idea of where you need to focus your work. And while yes, every little bit of performance you can squeeze out of a racecar is important, you also have to keep scale in mind. Let's say for example I know I can get an order of magnitude improvement in drag... is that 100 lbf versus 10 lbf? 10 lbf to 1 lbf? Or 1 lbf to 1.6 oz? When you can easily sidetrack yourself working on a million things at once, you've got to go after the low hanging fruit first, and really ask yourself, "What is the most efficient use of my time?"

Anyway, onward. Since aerodynamics aren't my strong point, you're all welcome to check my math.

When fabricating control arms and pushrods you've basically got two choices for tube profile. Round (as I've shown so far in CAD for simplicity's sake) and streamline. This particular streamline tube profile is available at Chassis Shop (they have some damn good pricing on CrMo tube!). While it's cool looking and tempting to use on everything there are also some definite constraints. It's generally 3-4x more expensive than round tube, it's more difficult to fit to itself for welding, and you have to make special end adapters for your rod-end and spherical bearings. This particular one I found on the Baumgartner Race Car page.

In order to justify the added expense, we should have a pretty damn good reason for using it.

Aerodynamic drag is generally given by the following equation:
Where F_d is the drag force, rho is fluid density, v is relative air speed, A is exposed frontal area, and C_d is the dimensionless drag coefficient. You had damn well better get your units straight for this one. Using Imperial units, density should be in slugs per unit volume.

C_d itself varies significantly with shape and with Reynolds number. For this particular analysis I'm assuming relatively high Reynolds number flow, where the drag on an infinitely long cylinder is around 0.5, and drag on a streamlined object is around 0.05.

I'm making a gross simplification and calling each control arm and pushrod a single tube exposed to the oncoming air. I'm also going to say they're all 0.625" diameter tubing, and all roughly 18" long. That gives me a grand total of 135 sq. in. frontal area. For velocity we'll use 100 mph (1760 in/s). For density we'll use 1.2 kg per cubic meter, which works out to 1.347e-6 slugs per cubic inch.

Turn the crank and I come up with 140.8 lbf drag. Given that it takes about 0.00267 horsepower to push 1 lbf at 1 mph (power = force * velocity)... at 100 mph that comes to 37.6 hp. That's quite a bit for a ~180 hp engine!

If I use streamline tube of the same height, since C_d goes down by a factor of 10, so does drag and power. 14.1 lbf drag, and 3.8 hp. Or in other words, going to streamline tube frees up 30+ hp at high speed. Looks like I'll be taking that approach. Realistically that might not mean an awful lot in terms of increasing drag-limited top speed since the tires and chassis have massive frontal area, but a couple extra mph makes a difference on a long straight.

Just for some perspective, if you were to do the same thing on a FSAE car... let's say at the top end of an acceleration run you're at around 60 mph. Let's assume we're using the same tube shapes and lengths. That's a reduction from 50.7 lbf to 5.1 lbf, and 8.1 hp to 0.8 hp. On a FSAE car then you'd expect to "free up" an extra 7 hp at high speed. On a restricted engine that's only making probably 75-80 hp to begin with, that's a non-trivial amount. To prove it to design judges of course you'd probably want to find a long stretch of asphalt and do acceleration, top speed and coast down runs with both styles of control arm. That'd be an interesting experiment and a good junior project.

6 comments:

ho said...

rho = 0.002378 slug/ft^3
Cd = 0.5
Area = 135/144 ft^2
V = 146.7 fts

Drag = 0.5*cd*rho*v^2*A

Drag = 12 lbf @ 100mph

-Chris Watson

Terps Racing

ho said...

I have no idea why my name is ho??

Jersey Tom said...

This is confusing the shit outta me now. I do come up with 12 with everything in feet.

In inches, I come up with 140. The unit conversion isn't exactly complicated.

Somethin's gettin screwed up..

sloracer said...

JT, your 140.8 is actually slug in/sec^2

need to divide that by 12 to get slug ft/sec^2... which is pounds on earth.

Unknown said...

what about round tubes with airfoil carbon covers

Joshua Gore said...

If your looking to keep costs down it seems something like thermoformed and bonded abs would be the cheapest way to make it. In black it could even look decent.