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Old 06-08-2007, 12:13 PM   #1
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Default Theoretical on G force elec on road

Any physicists out there with some math ability?

Slow day at the office today and my colleagues and I are having a discussion about 1/10 TC and what sort of G forces are developing as the car accelerates/brakes and takes corners.

Assuming you had a 1/10th scale person in the car what effects would these g-levels have on the person other than presumably a horrible death?

As I said slow day in the office...........
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Old 06-08-2007, 12:33 PM   #2
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Old 06-08-2007, 12:58 PM   #3
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The g-force is going to be the same for a 1/10 scale person or a full scale person (if you could fit him into a 1/10 scale car) :P
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Old 06-08-2007, 02:12 PM   #4
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Originally Posted by GCracker View Post
Any physicists out there with some math ability?

Slow day at the office today and my colleagues and I are having a discussion about 1/10 TC and what sort of G forces are developing as the car accelerates/brakes and takes corners.

Assuming you had a 1/10th scale person in the car what effects would these g-levels have on the person other than presumably a horrible death?

As I said slow day in the office...........
Hereís my back-of-the-envelope answer. If youíre interested in lateral acceleration, and you assume that the car is traveling at a constant speed through a constant radius turn, you only need the equation for centripetal acceleration (constant velocity circular motion): (velocity) * (velocity) / (radius). You can use whatever values you feel comfortable with to compute an answer. For example:

Lane width of typical indoor track = 10 feet = 3.048 meters
Assumed constant radius of turn (perhaps a 180 degree turn heading onto the straightaway) = 6 feet = 1.83 meters
Assumed constant velocity of stock motor TC navigating this turn = 15 mph = 6.71 meters / second
Acceleration of gravity = 9.81 meters / second^2

These values yield a constant acceleration through the turn of 24.6 meters / second^2

Expressing that value in multiples of the carís weight, you divide by the acceleration of gravity, yielding ~2.5g forces. Different assumptions will give different results. As far as damage to humans, I believe human factor engineers typically use 3g forces as an upper limit for tolerable acceleration, say when designing a roller coaster. That might be incorrect, though.

Andy
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Old 06-08-2007, 02:18 PM   #5
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I'd like to know what kind of g forces these cars pull under acceleration. At Reedy, the invite guys are getting from maybe 10 mph coming through the esses to 50+ at the end of the straight less than 120 feet away. Anyone know the math to figure that out? It's been 13 years since I was in physics class and can't remember that stuff.
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Old 06-08-2007, 02:31 PM   #6
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I'd like to know what kind of g forces these cars pull under acceleration. At Reedy, the invite guys are getting from maybe 10 mph coming through the esses to 50+ at the end of the straight less than 120 feet away. Anyone know the math to figure that out? It's been 13 years since I was in physics class and can't remember that stuff.
John,

You can compute average acceleration using the relationship:

(ending velocity Ė starting velocity) / (time interval)

The guestimating begins by making an assumption about when the cars hit their top speed of 50 mph. If we assume they do it half way down the 120 foot straightaway, this would yield:

Initial velocity = 10 mph = 4.47 meters / second
Final velocity @ end of 120 feet = 50 mph = 22.35 meters / second
Average velocity over first 60 feet = (10 mph + 50 mph) / 2 = 30 mph = 13.41 meters / second
Distance of half straightaway = 60 feet = 18.29 meters
Time to cover first 60 feet = 18.29 meters / (13.41 meters / second) = 1.36 seconds
Average acceleration = (22.35 m/s Ė 4.47 m/s) / 1.36 seconds = 13.12 meters / second^2

Dividing by 9.81 gives an average straight line acceleration of ~1.33g The peak, instantaneous acceleration is obviously more, but I'm not sure how much more.

Andy (I guess Iím having a slow day at work too - ha!)
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Old 06-08-2007, 02:49 PM   #7
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(Velocity^2)/radius = a

velocity in meters/second
radius in meters
aceleration in meters/second^2

divide acceleration by 9.8 m/s^2 to get G force.

6.71^2 / 1.83 = 24.60 m/s^2 = 2.51 G's. I agree with Adavid.


I would assume that the acceleration that occurs going down the straight will be higher because the cars reach top speed well before the end of the straight and they don't acclerate at a constant rate. The maximum accleration will probably occur right as the driver gets on the gas comming onto the straight. We would need a strobe like and a camera that could have its shutter left open. Then almost everything could be found based on the distance traveled and the time inbetween the strobes of light.

yes I am a little bored at work today too!!
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Old 06-08-2007, 02:58 PM   #8
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You guys are bored at work, I've been home from work sick all week. I'm bored as hell!
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Old 06-08-2007, 03:12 PM   #9
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Hey I just thought of this, I can measure the accleration of my TC5 using my eagle tree data recorder. It may not have good enough resolution at 10 samples per second, but I might be able to get an idea atleast. Hmmm may have to try that.
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Old 06-08-2007, 03:15 PM   #10
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Ok now how would that pan out using pro mod brushless set up
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Old 06-08-2007, 03:31 PM   #11
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Ok now how would that pan out using pro mod brushless set up
The average linear acceleration estimate of ~1.3g's was calculated using estimated speeds for the invite class at the Reedy race, so I think that would apply directly for a pro mod brushless setup. To me, the lateral acceleration is a function of the racing surface, the tires, and the ability of a well setup touring chassis to carry corner speed through a turn under those track conditions (and not the car's motor). You could vary the turn radius or the assumed constant corner speed to get a much different answer than 2.5g's lateral, but that seems (IMO) to have little to do with the car's straight-line acceleration or top speed.

Andy
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Old 06-08-2007, 06:43 PM   #12
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I agree with the 2.5g lateral... because it is possible to hold a constant speed while turning. a = v^2/r will work in this case.

However, i do not agree with the 1.3g, because this assumes the car undergoes constant acceleration on the straight... which i do not think is true... despite everything being linear... i think the frictional forces between the rubber tires and the road cannot provide the car with constant acceleration.

The "average" measurement does not reflect the change in acceleration enough... for the answer to be valid. But hey... i am only a highschool teacher, what do i know about real physics?
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Old 06-08-2007, 07:51 PM   #13
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I have measured 2.8 g's on indoor asphalt, rubber tires, medium radius turn, with a touring car. I used the Physics formulas similar to those posted above but time is measured instead of speed. 2.8 g's is similar to levels achieved by fullsize cars that have good ground effects systems on high speed turns. That would include GTP cars and F1 cars.

A traction roll is possible on asphalt. Especially during a long race as traction come up. A second formula for weight transfer can be used to find out at what point all the weight will transfer to the outside tires. At this point a roll over can occur. A slightly higher g level of 3.5 results from this set of calculations.

My pan car accelerates about 180 ft in about 1 second reaching 60 mph. Thats about 3 g's. Novak 3.5R, 2s LiPo. 40 ounce total weight. This is as good or a little better than an 1/8 scale Nitro spec class car. That acceleration time is not measured. The cornering time and radius were measured.

So here is a procedure on cornering g's. Find a medium corner and have a good driver tell you where the race line is. Corner entry and Corner exit points. Use a string and stretch it out until one end on the race line touches the corner entry point and when swung around the corner exit points. The other end is held stationary and the center of curvature of the turn. It helps to select a 1/4 circle section of the turn to time. Measure the string distance. Time the car several times on the 1/4 circle while the good driver is driving laps. I'll attach the calculations in a minute.
Attached Thumbnails
Theoretical on G force elec on road-cornering-power003.jpg  

Last edited by John Stranahan; 06-08-2007 at 08:41 PM.
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Old 06-08-2007, 10:08 PM   #14
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Ok, I looked at my eagle tree data and did some work in Excel to find a maximum acceleration of 3.27 G's. I did (V2-V1)/(T2-T1). This was achieved with a TC5 running 6 cells, a non_roar brushless motor and rubber tires. Kind of interesting. According to my data, my max G's was decelerating, -3.47 G's.
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Old 06-08-2007, 10:32 PM   #15
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Here is a little picture of my data, I appologize for the mistake in the acceleration units. It should have been m/s^2
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