given: the rate at which any object is accelerated on the surface of the earth is 9.8 meters per second, per second.

now heres where i may need help.

1.0G = 9.8m/s/s

how can you explain what 1G of lateral force is to someone in more relative terms?

is it like freefalling?

is weight a factor even though earth's force is irrelevant of weight? i.e. the force it takes to move a 100lb person sideways at 9.8m/s/s?

on the skidpad, if a vehicle is producing 1.0G, is the speedometer holding steady at say, 55mph because its going in a circle? must it be steadily increasing if going straight?

1.0G = XX amount of torque it takes to accelerate a 2000lb car roughly 13.0 seconds in the quarter (under ideal conditions)?

the force of being completely stopped from 30mph to zero?

is the force of the ground holding us in one place 9.8m/s/s against the gravity of the earth?

if so, 1.0G = the force it would take to hold your own body weight atop your head?

 

"1G" is equal to the rate of acceleration due to gravity on the earths surface.

9.8m/s/s

the amount of force that is required to accelerate the object downward depends on the mass of the object and is measured in newtons.

Force = mass x acceleration

9.8 Newtons = 1kg x 9.8m/s/s

the earths gravitational force IS dependent on the objects weight... well, more accurately, the mass.

the "weight" of the object is how we measure the force of gravity on an object.

 

for example, we know that two objects of different weights accelerate downward at the same speed, if you ignore air resistance.

but the force of gravity on them is different. Which is obvious, because it takes a lot more force to accelerate a car at a given rate then your body.

So, if you jumped out of a C-130 at the same time someone drove your car out of it, gravity would be pulling a lot harder on the car, even though you're accelerating at the same speed.

remember, force = mass x acceleration

YOU:
980 Newtons of force = 100kg x 9.8m/s/s

YOUR CAR:
11760 Newtons of force = 1200kg x 9.8m/s/s

you are both being accelerated downard at the same rate, but gravity is pulling a lot harder on your car.

what I'm trying to say is "1G" is a rate of acceleration, not a specific amount of force.

1G is the same acceleration for a 5,000 H2 as it is for a geo metro. They'll both cross the 1/4 mile marker in about 9 seconds. Theres just a helluva lot more force needed to make the H2 accelerate at the rate of what we know as 1G.


in answer to your skidpad question, although the speedometer will stay the same because you aren't going faster, there is a constant acceleration vector perpendicular to that of your path over the ground, which makes you turn around in a circle.

 

Check please.

 

A g is simply an acceleration constant = 9.81 m/s^2 (32.2 ft/s^2). This is used as an approximation for the acceleration due to the gravity of the earth. It is not limited to gravity. You can describe the acceleration in g's in any direction.

You seem to be a bit confused on some physics though. An acceleration isn't a force. It's a change in velocity with respect to time (if you take some upper level physics classes, you'll describe acceleration through a material derivative of a velocity field but don't worry about that).

About the skidpad, the speedometer doesn't change because it displays speed. Your direction is changing as you go around the circle, however, so your velocity is changing. After a bit of time, a car is moving at the same speed on the skidpad (we'll say 55mph). For just circular motion (which you are pretty much doing) the acceleration of the car is a = v^2/R. v is your speed, and r is the radius of the skidpad. This acceleration is the result of the tire forces on the road. So you aren't turning because you are accelerating. You are accelerating because you are turning. This kinematic (don't care about forces) approach can be confusing when you start kinetics (determine accel from forces).

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by GSpeedR &raquo;</TD></TR><TR><TD CLASS="quote">A g is simply an acceleration constant = 9.81 m/s^2 (32.2 ft/s^2). This is used as an approximation for the acceleration due to the gravity of the earth. It is not limited to gravity. You can describe the acceleration in g's in any direction.

You seem to be a bit confused on some physics though. An acceleration isn't a force. It's a change in velocity with respect to time (if you take some upper level physics classes, you'll describe acceleration through a material derivative of a velocity field but don't worry about that).

About the skidpad, the speedometer doesn't change because it displays speed. Your direction is changing as you go around the circle, however, so your velocity is changing. After a bit of time, a car is moving at the same speed on the skidpad (we'll say 55mph). For just circular motion (which you are pretty much doing) the acceleration of the car is a = v^2/R. v is your speed, and r is the radius of the skidpad. This acceleration is the result of the tire forces on the road. So you aren't turning because you are accelerating. You are accelerating because you are turning. This kinematic (don't care about forces) approach can be confusing when you start kinetics (determine accel from forces). </TD></TR></TABLE>


when someone then talks about a 1.0 g skidpad reading, what is being described? is the 1.0g's a description of the centripetal acceleration?

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote &raquo;</TD></TR><TR><TD CLASS="quote">Your direction is changing as you go around the circle, however, so your velocity is changing.</TD></TR></TABLE>

just to clarify for those who may be confused velocity is a vector (speed/magnitude and direction dependent) whereas speed is simply the absolute value of the magnitude of the velocity vector.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by bad-monkey &raquo;</TD></TR><TR><TD CLASS="quote">when someone then talks about a 1.0 g skidpad reading, what is being described? is the 1.0g's a description of the centripetal acceleration?</TD></TR></TABLE>

Yep. It's a measure of handling. Essentially, how much force can be applied to the car. Nobody wants to calculate the actual force, so we measure the acceleration due to that force.

 

i believe basically, say if u have a hanging ball on the rear view mirror, at 1G, the ball will be horizontal and parallel to the ground, if i remember correctly, i forget,

 

ok ya i think my mistake was not identifying 1G as an acceleration, which is a factor of force with mass being the other factor. so whether a hummer or civic, the person inside is experiencing the same force at 1G. the engine/tires of the hummer however, need to provide a much greater force to reach the 1G than the civic.

lets say a person is 150lbs, or 68kg. force = mass x acceleration,
so 68kg x 9.8m/s/s = 666 Newtons.
so is ~666 Newtons (or some other common occurance that produces ~666N) the best way to describe the force of 1G to someone?

 

This is a classic teaching situation that has trouble making the distinction between telling someone who doesn't know what 1g is, showing them how to do the math, and describing it so that layperson can actually internalize an understanding of it. I taught junior high school science and took college physics and I wouldn't have a clue what 666 Newtons feels like.

Your bathroom scale measures "weight" in pounds, which is an indication of the force (F) toward the center of the earth, which is the product of (or equals, = ) all of the molecules in your body (mass, m) being accelerated (a) at 1g.

(Ignore the fact that we are mixing units here, using pounds rather than Newtons. It does NOT matter for this kind of understanding.)

The same math works on, say, just your head. It has its own weight, the product of its own mass times that same acceleration of 1g.

Now, if that same head is being accelerated laterally - to the side - by the same amount (1g), it "weighs" the same amount it that direction as it does toward the center of the earth, the direction we normally think of "weight" acting.

If you could put that bathroom scale inside the driver's side window, stick your head up there, and turn right such that you got 1g of acceleration, the scale would read exactly what it would if you pulled your head off and plopped it on there in the bathroom.

Have the person you are explaining this to lie on their side on the floor, and hold their head up. Their noggin is experiencing 1g of acceleration in the direction that they commonly think of as sideways (ie. out their ear). In a car in the real world, that "sideways weight" (force) is added to the toward-the-center-of-the-earth weight because the latter doesn't go away.

If you have a CD or other sparkly thing hanging from your rearview mirror, it WON'T point straight out to the side when you hit 1g - it will be at a 45* angle. Your homework is to figure out a way to explain why without using the word "Newton."

K

 

i know!!

1g down + 1g to the right = a triangle who's angle is 45*

 

along the lines of what Herr Doktor Professor said, it might be useful to show that for a car on a certain diameter skidpad, 1.0g equals some particular speed indicated on ye olde speedometer.

g = (r in feet * 1.225) / t^2

so on a 100' radius skidpad, to have 1.0g lateral accelleration, you have to be traveling at a steady 38.7MPH.

did this help? or did I just confuse the issue further?

edit: sorry about the english units. I'm an old fart and don't think in metric

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by hondan00b &raquo;</TD></TR><TR><TD CLASS="quote">1g down + 1g to the right = a triangle who's angle is 45*</TD></TR></TABLE>

Now draw us the free body diagram!

 

I think G-tech makes a product for this discussion
you're fuzzy dice should be at 45 degrees assuming you are maintaining 1 lateral g
and you are on earth and you are measuring the angle relative to flat ground.
your car should be leaning some at 1g ..
so depending on how stiff your car is, the angle of the track, the ball should be around 45 degrees maybe a little less

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by Crazydave &raquo;</TD></TR><TR><TD CLASS="quote">I think G-tech makes a product for this discussion
you're fuzzy dice should be at 45 degrees assuming you are maintaining 1 lateral g
and you are on earth and you are measuring the angle relative to flat ground.
your car should be leaning some at 1g ..
so depending on how stiff your car is, the angle of the track, the ball should be around 45 degrees maybe a little less</TD></TR></TABLE>

howd you know! i just got a g-tech and on my first run got 0.95G's@55mph.

 

Let's pretend you've got a yo-yo. You are spinning it around your head. It has a certain mass, depending upon which one you bought. It is traveling in a circle, and the tension in the string, a force, is keeping it from flying away. The tension in the string is like the lateral grip from the tires for a car traveling in a circle. It is what keeps your car from flying away in a corner. The centripetal force F1=mv^2/r where r is the radius, with the centripetal acceleration being v^2/r. Also, a body in a gravitational field on the surface of the earth feels a force of F2=mg where g is 9.81 m/s^2. Thus the acceleration you feel, the one pushing your body towards the outside in the turn, depends on the speed squared divided by the radius of the turn. If you want to use g's as your units of acceleration instead of m/s^2, then just divide a or v^2/r by 9.8 m/s^2. Thus you have:

a(g's) = (1/9.81)*(V^2/r)

for a lateral acceleration of 1 g, you get a(g's) = 1 = (1/9.81)*(v^2/r)

An example:

If you take a 200 ft diameter skid pad, the radius is 100 ft. Multiply by 0.3048 to get radius in meters, which gives r=30.48 meters. then 1 = (1/9.81)*(v^2/30.48)

Re-arranging, you get

v^2 = 30.48*9.81 = 299.0

or v = sqrt(r * 9.81* a(g's)) = sqrt(30.48*9.81) = sqrt(299.0) = 17.29 meters/s

convert to v(ft/s) by dividing v(m/s) by 0.3048, thus v(ft/s) = 17.29/.3048 = 56.73 ft/s


to get mph, just multiply v(ft/s) by 3600 and divide by 5280

v(mph) = v(ft/s)*3600/5280 = 56.73*3600/5280 = 38.68 mph
(Note: it is sometimes useful to remember that 15 mph = 22 ft/s or 30 mph = 44 ft/s, etc.)

So if you have a constant speed of 38.68 mph on a 100 ft radius skid-pad, your lateral acceleration is 1 g. Lateral acceleration is very sensitive to the speed due to the square relationship. If you double the speed, your g's go up by 4 times, so there is no way you are going to double your speed in a corner, as the tires could never do it.

It is easier to measure time precisely on a skid-pad than speed, so if you want to know your acceleration in g's compared to time around the skid-pad, we have to compute the time for one lap of the circle.

t=distance/velocity and for a circle, distance = 2*pi*r, so t=2*pi*r/v, or v=2*pi*r/t

since the acceleration in g's is given by a(g's) = (1/9.81)*(v^2/r), if we plug in out value for g using time t, we get (with a little math)

a(g's) = (4*pi^2/9.81)*r/t^2 = 4.024*r/t^2 (for metric units)

or if you want to know t, then

t = 2.006*sqrt(r/a(g's)) (for metric units)

Back to our example of a 100 ft radius or 30.48 m skid-pad

if your speed is 38.68 mph or 17.29 m/s, we know we have 1 g of acceleration, and the time will be

t = 2.006*sqrt(30.48/1) = 11.07 seconds. Checking back for a(g's) if we knew the time,

a(g's) = 4.024*r/t^2 = 4.024*30.48/11.07^2 = 1.00 g's as expected.

Thus if you travel the skid-pad in 10 seconds, a(g's) = 4.024*30.48/10^2 = 1.227 g's (not too likely on street tires) and the speed is just v = 2*pi*r/t or 19.15 m/s or 62.83 ft/s or 42.84 mph.

If you run the skid-pad in 14 seconds, your g's are = 4.024*30.48/14^2 = 0.625 g's
If you run the skid-pad in 13 seconds, your g's are = 4.024*30.48/13^2 = 0.726 g's
If you run the skid-pad in 12 seconds, your g's are = 4.024*30.48/12^2 = 0.852 g's
If you run the skid-pad in 11 seconds, your g's are = 4.024*30.48/11^2 = 1.014 g's
If you run the skid-pad in 10 seconds, your g's are = 4.024*30.48/10^2 = 1.227 g's
If you run the skid-pad in 09 seconds, your g's are = 4.024*30.48/09^2 = 1.514 g's

All this to say that is you want your g's to be precise, you had better be measuring with a digital timer and not a stop watch, or else count a few laps and average the time. Also say you read a test about your favorite car in some magazine that states the max g's for it as 0.852 g's, and then you check some other car which has 0.870 g's. It sounds a lot better to you, but what does this mean? Time for 0.852 g's is 12.00 seconds from above and t for 0.870 g's is found from t = 2.006*sqrt(30.48/0.870) = 11.87 seconds, a difference of only 0.13 seconds which is a pretty easy error to make. Also how close to the center of the skidpad are you driving? You had better have a painted circle and a good driver.

A simple formula to use for imperial units is a(g's) = 1.227*r(ft)/t(s)^2

If we check, for a 100 ft radius and a time of 11.07 seconds, a(g's) = 1.227*100/11.07^2 = 1.00 g's as before.

Also in imperial units, v(ft/s) = 5.673*sqrt[(r(ft)*a(g's)], if r=100 ft at 1g, we get v=56.73 ft/s as before.

For an answer in mph, v(mph) = (15/22)*v(ft/s) = 3.868*sqrt[r(ft)*a(g's)], if r=100 ft at 1 g, we get v=38.68 mph as before.

A little physics goes a long way when you are talking about g's!

 

DOH! I didn't see where it said you had to show your work. Do I still get credit for answering the question? How about partial credit?

 

Sure. you can have all the credit! I'll keep the g's.

 

descartesfool, you lie!

anywho, with such a large margin of error with timing, what do you think about the g-tech and its use of an accelerometer?

 

I read this whole thread.. every word... even the stuff I don't understand!

&lt;--realizes that math/physics will be important in his life someday very soon.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by XDEep &raquo;</TD></TR><TR><TD CLASS="quote">
anywho, with such a large margin of error with timing, what do you think about the g-tech and its use of an accelerometer?</TD></TR></TABLE>

I had one of the original G-tech Pro's. It was fun for about a day. It was nice for acceleration times and to find you peak g's. However once you know how fast you can accelerate to 60 mph, there is not much fun it doing it over and over. I do not think it is a very useful device for the long term. I never used mine and sold it. That is why I bought a (relatively cheap) data logger from Race Technology. This charts your g's for a whole lap, whether around a skid-pad, a parking lot or a track. It shows the difference between peak g's and the average maximum sustained g's, which is what you are interested in, and what is measured on a skid-pad. There are always peaks in the lateral g chart, and you tend to ignore those, because they are due to things like tires catching on small imperfections in the pavement and momentarily increasing lateral grip. With the G-tech, I seemed to not really know what I was measuring, peak or average g's, and without the chart, you don't really know what is happening. There is no end of looking at data logger charts for lapping and setting up your car and driving. Using a stopwatch to time a few laps around a skid-pad would give you a much more accurate idea of both your g's and how the car was handling. The only problem is that most parking lot owners seem to balk when you ask them if you can paint a 200 ft diameter circle around the lamppost. You might have to tell them it is for some weird religion.

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by descartesfool &raquo;</TD></TR><TR><TD CLASS="quote">
The only problem is that most parking lot owners seem to balk when you ask them if you can paint a 200 ft diameter circle around the lamppost. You might have to tell them it is for some weird religion.</TD></TR></TABLE>

hah or you get a ticket like me have any of you actually been able to make use of someone's parking lot with the owner's permission?

 

HAHA, thats funny

Very good explination btw.

 

for physics and math

 

<TABLE WIDTH="90%" CELLSPACING=0 CELLPADDING=0 ALIGN=CENTER><TR><TD>Quote, originally posted by igyloo &raquo;</TD></TR><TR><TD CLASS="quote"> for physics and math</TD></TR></TABLE>

word!

some hard ****!

but pays off in the end when you learn it.


UCI student