blown 87

No, Mark, never have done it on final or close to the ground, had a very good friend do it once when a seat track broke.

My point is was I wanted to know if you had any knowledge of flying, you have answered that question in your usual way.

mark kibort

Greg, R

Regarding relative wind, I think you might be the one that is confused. The angle of attack IS the cord line vs the angle difference to the relative wind. Hey, i know what you are trying to say and I think you know what I was saying as well.
Basic stuff.
mk

blown 87

Been a very long time for me to remember stuff like that Mark.
That is why I said if I remember right.

I just looked it up on Wiki, copy and paste.

The angle between the chord line of an airfoil and the relative wind defines the angle of attack. The relative wind is of great importance to pilots because exceeding the critical angle of attack will result in a stall, regardless of airspeed.

and yes we both know what each other is trying to say, but you are saying that you will never see a true 0 airspeed and that is where we do not agree.
It can and does happen, but only for a fraction of a second.

My biggest problem is getting my feet to work when they should, I just have not done enough acro to have it be second nature and most likely never will.

Some things are easy for some folks, but I can not do a good 8 point roll to save my life.

BTW, the Ted Smith Aerostar is one of the best looking twins ever built.

mark kibort

I know, i was nit picking!

I just did a google and just found that too.

I never said you willl never see "0" air speed, I said you will never see "0" air speed and maintain altitude, and on top of that, you cant see 0 altituded and have 0 ground speed, with 0 windspeed. of course you can see 0 airspeed, if you are falling or starting to fall.

accelerated stalls. very dangerous. (stall, even with any air speed if the angle of attack is too great) . thats what kills a lot of folks in aerostars! thin wings, fast planes! Like flying race cars!!

Im with you on rudder input, but that is probably just lack of experience, (not good doing slip and crab landings, though I cant coordinate with model airplanes either, and ive been flying them longer.

mk

Richter12x2

I get it, so you're saying, dyno on donut tires!!!

mark kibort

Someone want to take a shot at plugging the numbers in?

24" diameter (keep it simple at 1ft radius), 50lbs (big 18" rims with 305 tires), assume even weight distribution.
Acceleration over 7 seconds from 1000rpm to 2000rpm. assume constant torque (rising hp with RPM).

This would show the total effects, measureable on the dyno in 4th gear with our cars on a dynojet, in the 300 hp range if you take that value X 2 for two wheels and tires.

Then, pull out 10lbs and see the savings.

Any takers?

James Bailey

that confirms what I had feared.....I am sure you always "WIN" there too

76FJ55

Okay assuming you are running along a flat torque curve with constant acceleration from 1000 wheel RPM to 2000 wheel RPM over a period of 7 seconds the following shold determine the amount of power consumed in accelerating the wheels at the 1000 and 2000 RPM points.

Equations:
Torque: T = Iα
α=angular acceleration (radians per sec^2)
I=moment of inertia of a disk = ½ M R^2
M=mass (lbm)
R=radius
Power: P(hp) = (T(lbf*ft)*RPM)/5252

With numbers:
α=1000RPM/7sec= 1000RPM*2πradians/60sec/7sec=14.96 rad/sec^2
M=mass in slugs=lbf*sec^2/ft
. 100 lb wheel: M = 100 lbm
. 80 lb wheel: f M = 80 lbm
I=moment of inertia of a disk = ½ M R^2
. 100 lb wheels: I=.5*100lbm*1ft^2=50 lbm*ft^2
. 80 lb wheels: I=.5*80lbm*1ft^2=40 lbm*ft^2
R=radius=1ft
Torque: T = Iα
. 100 lb wheels: T = Iα =50 lbm*ft^2*14.96 rad/sec^2 = 748.0 lbm*ft^2 /sec^2
. 80 lb wheels: T = Iα =40 lbm*ft^2*14.96 rad/sec^2 = 598.4 lbm*ft^2 /sec^2
. Converting lbm to lbf
. 100 lb wheel: 748.0 lbm*ft^2 /sec^2 * sec^2/32.2ft=23.23 lbf*ft
. 80 lb wheel: 598.4 lbm*ft^2 /sec^2 * sec^2/32.2ft=18.58 lbf*ft
Power: P(hp) = (T(lbf*ft)*RPM)/5252
Power at 1000 RPM used to accelerate the wheels:
. 100 lb wheels: 23.23 lbf*ft*1000/5252=4.427 hp
. 80 lb wheels: 18.58 lbf*ft*1000/5252=3.538 hp
Power at 2000 RPM used to accelerate the wheels:
. 100 lb wheels: 23.23 lbf*ft*2000/5252= 8.846 hp
. 80 lb wheels: 18.58 lbf*ft*2000/5252=7.077 hp

mark kibort

I always LEARN, there too, not always win. actually, we all win!

mark kibort

Excellent! A+ (extra credit for using both wheels as one)

so folks. copy this and show all your friends at dynos, over dinner, on the list. If you ever hear of any strange HP values gained by saving 20lbs of wheel weight, even though it was rotating, here is the HP gained on the dyno in probably near 4th gear. Now, these values will go up in lower gears, but on the dyno this is what you would see. 1-2 Hp! If you have more than this kind of gain on the dyno, its due to something else.

Thanks for doing and showing the work on that.

as a note, if you did have constant acceleration as you used, the engine actually would have a slight up tick in torque, without the wheels, right? But, since we are talking about dynoing, it really doesnt matter, because that is what the "to the wheels " forces would be and would register

Thanks again,


Mark

GlenL

I think there's something wrong here, or some unstated assumptions.

Power is consumed by the wheels accelerating. I don't see how fast the wheels are accelerating. Also, have you converted from Lbsf to Lbsm or slugs?

It might be easier to compute the total energy in the spinning wheel at, say, 150MPH and divide back by total time to get an average power consumed.

mark kibort

14.96 radians/sec^2 is the acceleraton rate

But this way, gives the HP at the rate of acceleration given, at 70mph and then at 140mph (approx speeds based on 1000rpm to 2000rpm wheel speed). It also assumes constant torque available for acceleration, in which case power will go up proportionally with speed.

I like this, because its a worst possible senareo. 7 seconds is MUCH faster than any of us would go from 70 to 140mph. also, its 10lbs per wheel and tire. even if you double the results for 4 wheels and tires, for net effect on the road, the HP effects are going to be sub 2 hp total, and MUCH less if a realistic acceleration is used. 400hp 928 might do 12 seconds from 70 to 140mph, so you can see the hp required for a much greater acceleration rate would be much lower. BUT, this discussion is the effects on a dyno. Over a 7 second run, the effects are in the example provided. we should have probably done a 3" , 25" diameter donut instead of a solid disc. That might triple the result values. In the end, that might be offset by the fact that our engines don't have flat torque curves to redline, so the acceleration rate would fall as RPM goes up, like a flat HP curve would. BUT, this example shown is just easier to understand and modify.

mk