Seat belt forces on car

  • #1
Hi all,

I'm having an online discussion and curious to know about the forces a 3-point seat belt has on a car.

One guy is fitting this retractable belt to his car.

P1000672.jpg


The retractor and anchor mount at one bolt hole, the anchor is off centre, it attaches via a steel extension piece, the section of extension piece you can't see is behind the retractor and bolted to the same hole the retractor is.

His car has two 'factory installed' lower mounting holes on one side, a separate hole each for the retractor and anchor and I advised him to mount the belt as it is here.

asa%20seat%20belt%20load%20chart%20-%20colored%20wording_thumb.jpg


Saying 'Personally I'd use the factory bolt hole, in an accident you have the force distributed between four anchor points not three'

I received a reply from someone saying

the force isn't divided up between four anchor points, it's really only three, the two floor ones on the side of the car and the one on the trans tunnel side.
The tunnel side gets 50% and the side ones 25% each


The top mount at the D-ring definitely experiences force so let's say you remove it.

Wouldn't the bottom anchor bolts feel an increase in force? When the person in the car experiences a sudden stop, doesn't the resultant force that the car experiences from the seat belt remain the same in both lap belt and lap-sash belt applications?

My opponent says adding an extra bolt will not distribute the forces evenly but I think he is assuming the tension T will be the same in both seat belt applications.

s
 
Last edited:

Answers and Replies

  • #2
35,667
12,228
To calculate tension in the belt, consider the required extension when the person moves a distance l: In the conventional setup, this is about 4l. Therefore, tension in the belt (integrated over the belt cross-section) is 1/4 of its stopping force F.
This is true for all belts which cross the body twice in some way.

To calculate forces on the anchors, simply count the number of belt parts connected to them: The outer anchor (same side as retractor) and the retractor will get F/4, the inner anchor and the D-ring get F/2.
 
  • #3
78
0
do it by parts mate, that should help
 
  • #4
consider the required extension when the person moves a distance l: In the conventional setup, this is about 4l.

Sorry but what do you mean by extension? The person in restraint has travelled distance L, does the belt stretch by 4 times the travel distance?

tension in the belt (integrated over the belt cross-section) is 1/4 of its stopping force F.

How did you work it out to be 1/4?

Is it because the person is held back at 4 points (shoulder, belt shoulder side hip, upper right hip, lower right hip) and at each point the force F= m/4 x a

Yes, Tension T = F/A (Cross sectional area)

This is true for all belts which cross the body twice in some way.

So the force in a lap belt (2-point belt) is twice at F = m/2 x a?

Or is there an elongation component to the equation and is the elongation more due to the restraining mass at only 2 points?

To calculate forces on the anchors, simply count the number of belt parts connected to them: The outer anchor (same side as retractor) and the retractor will get F/4, the inner anchor and the D-ring get F/2.

Wouldn't the force at the D-ring be 2FCos θ/2 ?

s
 
  • #5
35,667
12,228
Sorry but what do you mean by extension? The person in restraint has travelled distance L, does the belt stretch by 4 times the travel distance?
Approximately. Depends on the precise geometry of the system.
How did you work it out to be 1/4?
Inverse of the factor 4.
Is it because the person is held back at 4 points (shoulder, belt shoulder side hip, upper right hip, lower right hip) and at each point the force F= m/4 x a
Like that, yes.
So the force in a lap belt (2-point belt) is twice at F = m/2 x a?
If the belt just goes from one side to the other: Yes.
Wouldn't the force at the D-ring be 2FCos θ/2 ?
Ah, you are right. But the angle is small, so I don't expect a big difference.
 

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