Calculating G Force from 250km/h to 80km/h in 80m - Chris

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SUMMARY

The forum discussion centers on calculating G Force experienced by a driver of a 250cc International Superkart decelerating from 250 km/h to 80 km/h over a distance of 80 meters. The calculation reveals that the acceleration is -27.02 m/s², resulting in a G Force of -2.75 Gs. Additionally, it is noted that to accurately measure G Force in a vehicle, parameters such as acceleration and time are essential, as constant speed does not produce G Force. The discussion also touches on practical applications, including the use of accelerometers for monitoring G Force in vehicles.

PREREQUISITES
  • Understanding of basic physics concepts, particularly acceleration and G Force
  • Familiarity with the formula V² = V₀² + 2aΔX
  • Knowledge of converting speed from km/h to m/s
  • Experience with accelerometers and data acquisition systems
NEXT STEPS
  • Research the formula for calculating G Force in different scenarios
  • Learn about accelerometers and their applications in automotive settings
  • Explore the impact of friction and wind resistance on G Force calculations
  • Investigate data acquisition systems like Motec for real-time G Force monitoring
USEFUL FOR

This discussion is beneficial for automotive engineers, physics students, race car drivers, and anyone interested in understanding the dynamics of G Force in vehicles.

ComposerCK
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Hi,

I'm hoping someone can provide me with a formula to calculate the amount of G Force experienced when decelerating, based on this information :

A 250cc International Superkart is traveling at 250km/h and is approaching a tight corner than can be taken at a maximum of 80km/h. The driver leaves the braking to the last possible moment, and decelerates from 250km/h to 80km/h in a distance of 80m (whilst in a straight line).

How many G's has the Superkart and driver experienced ?

Thanks, Chris.
 
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It's pretty straight forward if you neglect friction and wind resistance, etc...

Note: 250 km/hr = 69.4 m/sec and 80 km/hr = 22.2 m/s

V^2 = {V_o}^2 + 2*a* \Delta X

(22.2 \frac{m}{sec})^2 = (69.4 \frac{m}{sec})^2 + 2 a (80 m)

-4323.5 \frac{m^2}{sec^2} = 160a m

a = -27.02 \frac{m}{sec^2}

g's = \frac{a}{9.81 \frac{m}{sec^2}}

g's = -2.75

I really hope this wasn't homework. If it was, you got a freebie.
 
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Thanks

Thanks for that !

No, it wasn't homework of any sort ... unfortunately I'm not studying Physics at all.

I must look into that formula so that I understand the parameters used and their relationship.

The speed into metres per second was obvious anyway, but I probably lack the basics to understand and agree with the rest of it. I can see that 4816.3 - 492.84 = 4323.52.

I wouldn't have thought friction and wind resistance would have been a factor in G force ... ?

-2.75 G's ... my guess before reading the answer was between 2 to 2.5 G's, based on what I've heard for a Formula One car ...

Once again, thanks for taking the time to reply ! :smile:
 
Hello.

I'm new on this forum so hello to everyone.

I'm attempting to monitor the G Force in a vehicle whiles it's driving, does anyone have any ideas?

All I have is the speed of the vehicle, would this be enough to measure the G Force?

jason
 
huttojb said:
All I have is the speed of the vehicle, would this be enough to measure the G Force?

jason

No. If the vehicle is moving in a st line with constant speed, there is no g force due to motion of the vehicle.

You have to know the acceleration. If you are moving with a constant speed on a circular track, then you have to know the radius of the track and the speed.
 
But obviously the vehicle needs to go from 0mph to (say) 90mph,

Obviously any variation in speed will result in G Force?? Is this correct? Let's say for argument sake, it's in s straight line?

How would I work out a running G Force? i.e. Like the one on F1 (ITV)
 
Lets expand on that.

If I made a trip from A to B, I would want to know the maximum amount of G Force I went through during this trip?

Can you Help.

Ta
 
ComposerCK said:
A 250cc International Superkart is traveling at 250km/h and is approaching a tight corner than can be taken at a maximum of 80km/h. The driver leaves the braking to the last possible moment, and decelerates from 250km/h to 80km/h in a distance of 80m (whilst in a straight line).

How many G's has the Superkart and driver experienced ?

Thanks, Chris.

The actual g force experienced by the driver would be the magnitude of the vector sum of 'a' backward and g downward, which will be equal to sqrt(a^2+g^2)=2.93g. This g force is at an angle tan-inverse(a/g) to the vertical.
 
huttojb said:
Lets expand on that.

If I made a trip from A to B, I would want to know the maximum amount of G Force I went through during this trip?

Can you Help.

Ta
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Hi Huttojb,

You need some more parameters to find the accn or the g force. If you know the time 't', and accn is constant and the motion is in a st line, then the accn 'a' is (90-0)/t mph/h.

If you know the dist, then too you can find the accn if it is constant.

In practice, the accn is generally not constant. So you would need a detailed dist-time graph; or hang a spring balance in the car to see the effective g force. (Note that the reading on the balance would include the g field of the earth.) The maximum accn would perhaps be when you start and reach a high speed in a few seconds. Or when you slow down by stepping on the brakes very hard.
 
  • #10
huttojb said:
I'm attempting to monitor the G Force in a vehicle whiles it's driving, does anyone have any ideas?
Do a web search for "accelerometer car", you'll find a few companies that sell these, or do a search for "data acquisition car" for high end systems such as this Motec:

http://www.motec.com/products/dash/adl2dash.htm
 
Last edited by a moderator:
  • #11
I have a Physics problem that I can't Figure out:

An Aircraft goes from 0 MPH to 85 MPH during its take off run of 800 feet in 10 seconds. The aircrafts take-off weight is 2600 pounds and has 4 people aboard with 40 mounds of bags. What is the G-force experienced by the People on board?









'
 
  • #12
we did a lot of posts on this on MEch engineering forum and other forum here

Centripetal force- Car making a turn Jun6-10, 06:57 AM

race car suspension class Jul22-09, 11:03 AM

see post below in gen pyshics forumn..
Wider tire giving better traction debate! Old Aug15-09, 04:19 AM
 

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