Team Camaro Tech banner

67 Installing 3 point seat belts in back seat

16K views 41 replies 18 participants last post by  morrisclassic  
#1 ·
Anyone have kids driving? I do. Wondering if an how to install 3 point belts in back seat. I would like two in front and two in back. Anyone got a suggestion?
 
#2 ·
I want to do this too. I spoke with billy from morris classic concepts and he didn't have anything yet, and don't know if he is working on it, but when I forwarded the pics of the original mounting (3rd) location on the package tray shelf, he said he had seen the pics but didn't know if that would be a strong enough/suitable location for them to anchor. I am hoping it is, but I think bugging him again about this couldn't hurt! So send him an email, or call and see if they have anything in the works! I think they would sell extremely well.
 
#6 ·
Installing 3 point seat belts in back seat

We are having mounting systems fabricated now. As with any project we
have to keep fitting and adjusting the template until we get the correct fit
and a safe product. The bracket will have three mounting points,4 bolts and will require the installer to drill a 1/2" hole.Then we will have a laser file made to cut the bracket. Then a metal fab shop to bend offsets in the bracket to fit properly. Billy
 
#8 ·
I started researching this last year and talked to one of the board members CarlC about it. Here is some info he passed me:


Carl, thanks for the pictures and information. I am going to try and
install them in the next month or two. I will probably email for more info
then. One question about putting the seat belt "loop" on the package tray.
Do you think it was necessary to strengthen the area up (where it bots to
the tray) or would a fender washer on both sides of the tray have been
enough?


Chris,

The loop attachment point has a much higher loading than the mechanism. Since the belt is pulling both forward and down the actual load is higher than the belt tension by a factor of 1.4 and pointing forward and down at a 45* angle. It might be fine with just washers but since there is no practical way to non-destructively test it I decided to be cautious and install the plate.




Here are some pictures he sent me:










I haven't started this but it is on my list of to do stuff (since I take my small children for rides in my car). BTW, I am currently using lap belts in the back seat.
 
#9 ·
interesting pics...

and wouldn't the load on the upper mounting point be sqrt2/2, not just sqrt2? as in .714, basically taking 71% of the load assuming a 45* angle? I don't think the mounting point load would be more than the actual load due to any angle for that matter? Does that make sense? The load would be distributed between the mounting points, but not necessarily evenly (all depends on relative angles with one another right). Also, there is actual torque force applied to the mounting stud or w/e that is on the package tray vs. if it was say hanging on the ceiling (not as much anyways).

All that asside it has to be strong. But, those belts look decent, although I would go for black!
 
#14 ·
Installing 3 point seat belts in back seat

Wendall,Thank You and Everyone for buying our products. Our goal is to make a Safe and Quality product. We make our products from quality materials and we have craftsman making our products that is some of the reasons our products are priced higher than other companies but we won't and can't compromise Safety and Quality. I'll keep everyone updated on the rear seat belts. Thank You, Billy
 
#15 ·
There is the forward force component (shoulder belt load) and a similar downward component (to the retractor mechanism) in my setup. Since the loadings are similar (assume them to be 1), then the resultant at the package tray belt guide mounting point will be the belt load * sqrt2 at a 45* downward angle.

The actual loading during an accident at the belt guide attachment point will be approximately the belt tension * sqrt2 as long as the right triangle arrangement remains in place. If the angle decreases, the loading increases. If the angle increases, the loading decreases.

The sqrt2/2 cannot be used to determine the loading at the belt guide point. The loading at the belt guide point can only be determined by the belt tension. The belt tension (under load) is calculated by figuring out what belt tension would be required to restrain a mass of X accelerated at Y with the belt at an angle Z (simple case). The sqrt2/2 cannot be assumed since the belt angles used to restrain the load are not usually 90*, and there may be more than 2, usually 3, and could be as many as 5 or more mounting points to share the load.

I bought the belts from seatbeltpros.com years ago, long before Billy came up with the kit. Given the amount of fabrication and welding I did, I'd go with his.
 
#17 ·
ha ha ha, I was waiting for someone to chime in here!

I still think I am right about this, and I know it would technically vary with the more mounting points and disperse more and more. But, your force vector is the body moving in basically a straight line forward... but because of the shoulder strap etc, the force gets absorbed at multiple points, and thus the force at any one point cannot exceed the total force in the system... it just can't. At least that makes perfect sense in my head!

I am not trying to give you a hard time, I was just trying to set up a basic force vector drawing and it wasn't clicking for me.

Bottom line is that the mounting points will need to be strong, which I think we are all in agreement.
 
#18 ·
The belt load is created by the body moving forward. No argument there.

So, instead of using 1 for a belt load, let's call it L. L is the same along the length of the belt during tension.

The load L of the belt goes up the shoulder and turns horizontally. The load L has not changed in scale, only direction. Now the belt load must turn downward through the guide. Assume the belt guide is a pulley (zero friction.) So now there is a horizontal and vertical belt load of L acting on the pulley. Draw a FBD of a right triangle with equal legs of L. The hypotenuse is R.

L^2 + L^2 = R^2.
sqrt(L^2 + L^2) = R
sqrt(2L^2) = R
L*sqrt(2) = R
Assume the load = 1
R = sqrt2, or approximately 1.414 at a 45* downward angle.

So, whatever the belt load is, the belt guide in my example/car sees a loading of 1.4L @ 45* down. I think what you are thinking are the loads at the belt ends. I wish it was easier to draw triangles on these reply sections....
 
#19 ·
Anyone have kids driving? I do. Wondering if an how to install....
Oh that brings back memories....
These now have their own children ...
Image
 
#20 ·
ah, but you see that when you apply the load to the shoulder strap (ie person going forward in a crash), you don't have an equal load "L" applied both downward and forward as you mention. If that were the case, then yes you would be right. But, as you mention, I speak about the dispersing of the load somewhat via the mounting bolt locations (in this case on the floor or w/e), which DO play a part in the equation. You can think of the shoulder point as a pulley, which is great because then you can visualize the load being transfered around the pulley down to the other 2 mounting bolts. But, in a crash, the retractor doesn't extend belt webbing, it is still and the "pulley" acts as a fixed point essentially, which really doesn't matter in this case, but stay with me!

Let me try to simplify my reasoning here... If you had one teather to the shoulder mounting location (in a one mount system), and applied a force of L to the teather say forward, then you would have a load of L on the shoulder loop location right?

So, adding 2 more mounting points down low just helps distribute the load which is a combined load (not necessarily evenly) of "L" between the multiple locations.

I am just saying that you can't gain force in a closed setup. if the force on the belt at one point is "L", then the force on any one mounting location on a multiple mounting location setup is going to be less than L.

Wouldn't you agree that if you had a 10 point harness, that if a load L is applied anywhere in the system that at any given point the load on any one mounting point or loop/pulley is not greater than the initial applied force?

It is also like picking up an engine with a chain that has 2 mounting locations. If your example were true, then you would be better with 1 mounting location, because in a 1 mount system the load would only be the total load, and not 1.414 times the original load. We both know that is not the case as 2 mounting locations are better and stronger than one mounting location of equal material composition, because the force on each mount is less than the original force or in this case weight of a motor.

Ok, I'll stop hijacking this thread for now! If you still don't agree with me, then we can agree to disagree! (even though I am right! aha hahha ahaha)-sorry, I had to!
 
#22 · (Edited)
Wouldn't you agree that if you had a 10 point harness, that if a load L is applied anywhere in the system that at any given point the load on any one mounting point or loop/pulley is not greater than the initial applied force?
I disagree, and I'll provide an example using an everyday three-point seatbelt system.

Note: A pulley can be used to both change the direction of tension and as a force multiplier. A belt guide, not belt endpoint, is basically a pulley with some friction.

Sit in the seat and pull the seatbelt over your chest. The tongue/male blade slides along the length of the belt via a guide that allows for ease of use and adjustability. The guide is in its simplest form a pulley. Let's ignore the friction component for this example since it is not needed to illustrate the basic reactions. The belt is now attached at two points.

One one end of the belt is the retractor mechanism, the other a hard mounting point.

Finish putting the seatbelt on by inserting the tongue into the latch. There are now three mounting points.

Pull up on the section of belt that is going across your chest so that the belt across your lap increases in tension. Now you have tension in the shoulder and lap belt passing through the guide in the tongue blade. If you pull on the belt as described above with 10#, then the lap belt also must have a 10# tension.

For ease of understanding assume that the shoulder and lap belt sections are parallel to each other as they enter the guide on the tongue blade. The tongue blade must therefore have a reaction of 20#. Hence, the reaction at the mounting point is 20#. The belt only has 10# of tension, but the guide bar, tongue, buckle, all the way down and including the bolt/floorboard must be designed to withstand a 20# load. This is a worst-case scenario.

The belts as they enter the guide on the tongue will typically not be parallel when inserted into the latch, but more than enough to create a reaction load far greater than the magnitude of the belt tension. At 90* to one another, as noted in the previous reply, is sqrt2.

I am just saying that you can't gain force in a closed setup. if the force on the belt at one point is "L", then the force on any one mounting location on a multiple mounting location setup is going to be less than L.
That is one of the primary reasons for using pulleys. Force multiplication. It's not what we want in some cases, but it's also how a single man can lift something dozens of times his own weight. It is also why the reaction at the tongue blade guide is significantly higher than the belt tension alone.

Some math:

A rope/belt is pulled horizontally at a tension T. In the middle of the belt (F) a frictionless pulley of negligible diameter is placed below the belt. Find the angle theta (0) to the horizontal where the vertical reaction (Fr) at F is the same as T. Assume that both ends of the belt are lowered an equal amount to the horizontal. (A drawing is worth a thousand words...)

Assume the downward direction is negative, upward positive. No horizontal movement occurs since both ends of the belt are dropped equally, hence the horizontal components remain equal and opposite.

Since we want to find the point where the belt tension is equal to the reaction Fr, then Fr=T.

For solving purposes, Fr=T=1

Using statics:

-Tsin(theta) - Tsin(theta) + Fr = 0
-2Tsin(theta) + Fr = 0
Tsin(theta) = Fr/2
sin(theta) = 1/2 {Fr = T = 1}
theta = 30*

Hence, whenever a minor/inside angle of less than 120* (180 - 2x30*) occurs then the reaction at the guide reaction point will be greater than the belt tension.

This is all a very simplistic engineering viewpoint. It does not take into account a multitude of other factors that come into play.

The reaction at the other end of the belt (opposite the latch) will be 10#, assuming there is no friction between the belt and your clothes.

Your never-greater-than-the-belt-tension reaction only works on a two-point lap belt type setup.
 
#28 ·
67-69 Installing 3 point seat belts in back seat

Hi Everyone,
We are sending the brackets to be tested this week. Then a laser file for cutting . The brackets will be made from stainless steel. Installation will be straight forward, Easy. All grade 8 bolts, washers and nuts will be included in the kit. We want the seat belt to break during the test so it will verify the strength of out brackets and assembly points. We don't want any doubt that our system will stay intact in the event any of us have to rely on them, which
we hope never happens. We are hoping to have them available within a month. Billy