How can I determine the center of mass of a vehicle?

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SUMMARY

The discussion focuses on determining the center of mass (COM) of a vehicle using various methods. The primary approach involves calculating the mass and positions of major components and using the reactions at the tires to find the COM. Participants suggest tipping the vehicle to measure the height of the COM and emphasize the importance of understanding weight distribution across the tires. The conversation highlights the necessity of using a weighing platform to obtain accurate readings of tire reactions, which are crucial for calculating the COM.

PREREQUISITES
  • Understanding of static equilibrium and forces in mechanics
  • Familiarity with Free Body Diagrams (FBD)
  • Knowledge of weight distribution principles
  • Experience with measuring equipment, specifically vehicle weighing platforms
NEXT STEPS
  • Research methods for calculating center of mass in rigid bodies
  • Learn about Free Body Diagram techniques for complex systems
  • Explore the use of vehicle weighing platforms and their calibration
  • Investigate the effects of vehicle tilt on weight distribution and COM calculations
USEFUL FOR

Mechanical engineers, automotive technicians, and students studying vehicle dynamics or mechanics who need to determine the center of mass for various vehicle types.

  • #31
Okay. I'll explain what I perceived from your posts.

That you said that what I need to do in order to find reactions at the tyres is simply to put the front two tyres and get their reactions and repeat this for the back two tyres. My point here is that the reading on the scales wouldn't be the 'real' or 'absolute' reaction of the tyres as the reading would change with the amount of portion of car over the platform with even only two tyres over it. So, the readings would change if we move the car around.
 
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  • #32
Altairs said:
My point here is that the reading on the scales wouldn't be the 'real' or 'absolute' reaction of the tyres as the reading would change with the amount of portion of car over the platform with even only two tyres over it. So, the readings would change if we move the car around.

Hi Altairs! :smile:

The reading doesn't change with the amount of portion of car over the platform, it only depends on which tyres are on the platform (and on the angle, of course).

If we put the front two tyres on the edge of the platform, and the rest of the car over solid level ground, the weight shown will be exactly the same as if we move the car forward so that the rear two tyres are almost on the platform. :smile:
 
  • #33
Then all I need to do is to weigh the front two tyres and divide it by two (assuming that they support equal weight) and do the same for the back two tyres.
 
  • #34
Altairs said:
Then all I need to do is to weigh the front two tyres and divide it by two (assuming that they support equal weight) and do the same for the back two tyres.

Yup! :smile:

(except why bother to divide by two … you only need the ratio … :wink:)
 
  • #35
When I'll take side view and front view than I'll only be considering the two tyres in view then won't I be assuming that the whole weight of the car is supported by those two tyres ?
 
  • #36
Hi Altairs! :smile:
Altairs said:
When I'll take side view and front view than I'll only be considering the two tyres in view then won't I be assuming that the whole weight of the car is supported by those two tyres ?

(I assume that by "view" you mean "on the platform")

Nooo … the total weight of the car is still supported by all four tyres.

You are measuring how much is supported by those two (and the rest of the total weight of course is supported by the other two, and you take the ratio). :smile:

(oh … but why take "side view" at all, if the c.o.m. is in the middle?)
 
  • #37
No with side view I mean the real side view. With the doors in view. By taking moment along one tyre and as I would know the reaction and the total mass of the whole car won't I be able to find the centre of mass in one axis ?

Doing the same for the front view (the windscreen) one I can find the COM in another axis ?

Is this feasible ?
 
  • #38
Altairs said:
No with side view I mean the real side view. With the doors in view. By taking moment along one tyre and as I would know the reaction and the total mass of the whole car won't I be able to find the centre of mass in one axis ?

Doing the same for the front view (the windscreen) one I can find the COM in another axis ?

Is this feasible ?

Yes, that's right … if you know the total weight, then you can find the x-coordinate of the c.o.m. by weighing the front tyres on the flat, the y-coordinate of the c.o.m. by weighing one set of side tyres on the flat, and the z-coordinate of the c.o.m. by weighing any pair of tyres on a slope. :smile:
 
  • #39
Right. Let's say that in the side view the tyres are T1 and T2. And in front view the tyres are T1 and T4. For calculating the coordinate in the side view I'll only need one of T1 and T2 and in the same way for calculating the coordinate in the front view I only need one of T1 and T4. I hope that I am right here.
 
  • #40
Altairs said:
Right. Let's say that in the side view the tyres are T1 and T2. And in front view the tyres are T1 and T4. For calculating the coordinate in the side view I'll only need one of T1 and T2 and in the same way for calculating the coordinate in the front view I only need one of T1 and T4. I hope that I am right here.

I'm confused … are you saying that you only need to put one tyre on the platform each time? :confused:

i] You can't put one tyre on, and then just double the weight;

ii] it's physically much easier to put two tyres on anyway, isn't it? :smile:
 
  • #41
I will be weighing two tyres at once.

My confusion still exists. I'll explain.

If there are two tyres over the platform (let's say front two) then changing the position of the car over the platform in such a way that rear two tyres are still on the ground should change the reading on the meter because the more the portion of the car is over the platform the more the reading will be. Is this the case ? You have already said both yes and no in previous posts.
 
  • #42
If there is the whole car on the platform and we lift it up at an agle such that it is just enough to put the rear two tyres in air and then take the reading. Then repeat this by tipping it on the back tyres then will the two readings equal the weight ?
 
  • #43
Altairs said:
If there are two tyres over the platform (let's say front two) then changing the position of the car over the platform in such a way that rear two tyres are still on the ground should change the reading on the meter because the more the portion of the car is over the platform the more the reading will be. Is this the case ? You have already said both yes and no in previous posts.

It's definitely no … the reading depends only on which tyres are on the platform (and the angle of the car), not on the "overlap".

I didn't intend to say "yes" … where do you think I did?
Altairs said:
If there is the whole car on the platform and we lift it up at an agle such that it is just enough to put the rear two tyres in air and then take the reading. Then repeat this by tipping it on the back tyres then will the two readings equal the weight ?

Do you mean lifting the rear tyres just an inch off the ground, or do you mean "balancing the car on its front legs"? :confused:
 
  • #44
tiny-tim said:
Do you mean lifting the rear tyres just an inch off the ground, or do you mean "balancing the car on its front legs"? :confused:

Just an inch. BTW what's the difference?
 
  • #45
tiny-tim said:
It's definitely no … the reading depends only on which tyres are on the platform (and the angle of the car), not on the "overlap".

It means that while the front two tyres are over the platform the reading will remain constant and it will shoot to the maximum value (weight) only when the previous two tyres are also over the platform. And in this duration (i.e. just before the rear two tyre are on the platform and just after the front two tyres are over platform) the reading won't change ?
 
  • #46
Altairs said:
Just an inch. BTW what's the difference?

Well, if you lift the rear tyres an inch off the ground, then that means that you are now supporting the same weight that the ground was supporting … so the platform reading (from the front tyres) will be the same.

But if the car is balanced (so that if you let it go, it would stay there), then all the car's weight is supported by the front tyres, and so the platform reading will show the whole weight. :smile:
 
  • #47
Here it is.

Altairs said:
But the problem is that let's say at one time only 1/4 th of the car is on the weighing platform with the front wheels.

Next time, half of the car is over the platform but still only the front two tires are over the platform. Won't the readings change ?

tiny-tim said:
Yes you can … if you put one or two tyres on the scale, and the others on the ground, then the scale will measure the weight that the scale is supporting.

By Newton's third law, that's equal and opposite to the (vertical component of the) reaction at that point! :smile:
 
  • #48
Altairs said:
It means that while the front two tyres are over the platform the reading will remain constant and it will shoot to the maximum value (weight) only when the previous two tyres are also over the platform.

Yes, that's right … drive the car slowly forward,and the platform reading will stay the same right until the rear tyres reach the platform. :smile:
 
  • #49
tiny-tim said:
Well, if you lift the rear tyres an inch off the ground, then that means that you are now supporting the same weight that the ground was supporting … so the platform reading (from the front tyres) will be the same.

But if the car is balanced (so that if you let it go, it would stay there), then all the car's weight is supported by the front tyres, and so the platform reading will show the whole weight. :smile:

The idea is that the whole car will be over the platform. Then the car will be lifted from it back by men or something so that they will be standing on the ground. My point was that there is no difference in this and simply putting front two tyres over and rear two tyres off the platform but my friends wouldn't agree. Second point is that will this procedure (the tipping one) when performed for both the front and the rear tyres equal the weight ?
 
  • #50
Altairs said:
Here it is.
Altairs said:
But the problem is that let's say at one time only 1/4 th of the car is on the weighing platform with the front wheels.

Next time, half of the car is over the platform but still only the front two tires are over the platform. Won't the readings change ?

Ah … now I see …

No, when you said 1/4 th of the car, I thought you meant that only one tyre was on the platform
 
  • #51
tiny-tim said:
Ah … now I see …

No, when you said 1/4 th of the car, I thought you meant that only one tyre was on the platform

Oh...right..
 
  • #52
Altairs said:
The idea is that the whole car will be over the platform. Then the car will be lifted from it back by men or something so that they will be standing on the ground. My point was that there is no difference in this and simply putting front two tyres over and rear two tyres off the platform but my friends wouldn't agree.

If the men are lifting it by holding onto the rear axle, then your friends are wrong, and you are right … the distribution of the forces (or reactions) depends only on the positions of the points of support on the car.

(But if the men are holding it by the tow-bar under the rear bumper, then the reading will change, because the c.o.m. is further from the tow-bar.)
Second point is that will this procedure (the tipping one) when performed for both the front and the rear tyres equal the weight ?

Sorry … I'm not following this at all … why would you get men to tip the car when you can get the same reading by leaving the rear tyres on the ground?
 
  • #53
when we tilt the car and find the reaction thn v'l have a frictional force acting on the tyre,in contact with the ground apart from the surface reaction so will we consider it as limiting friction or how do we calculate it?
 
  • #54
Welcome to PF!

Whatevr said:
when we tilt the car and find the reaction thn v'l have a frictional force acting on the tyre,in contact with the ground apart from the surface reaction so will we consider it as limiting friction or how do we calculate it?

Hi Whatevr! Welcome to PF! :smile:

The weighing platform will only measure the vertical reaction, and the friction force is entirely horizontal, so the friction force will make no difference. :smile:
 
  • #55
but when u find the z distance u take moments abt a point nd therefore u have 2 kno all the forces how else wud u claculate the distances
 
  • #56
Whatevr said:
but when u find the z distance u take moments abt a point nd therefore u have 2 kno all the forces how else wud u claculate the distances

Hi w'v'r! :smile:

ah, you've made a good point … if the higher pair of tyres is supported on a sloping surface, then the friction will spoil the equations.

So it is important that the higher pair of tyres (or its axle) be supported on a raised horizontal surface. :smile:
 
  • #57
no i don't get it:( that won't b possible i thnk or the frictional component wud stil exist ??
friction will exist if we raise the car frm one end
 
  • #58
Whatevr said:
no i don't get it:( that won't b possible i thnk or the frictional component wud stil exist ??
friction will exist if we raise the car frm one end

If all the wheels are on horizontal surfaces, why would the car want to move anywhere (even with the brakes off)?

There'd be no gravitational advantage in moving. :smile:
 
  • #59
no but thn how'd we calculate the z distance i.e. the height part?
 
  • #60
Whatevr said:
no but thn how'd we calculate the z distance i.e. the height part?

If we can do away with the friction part than it is quite simple.
 

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