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Thread: Finding Tongue Weight on a Trailer with Multiple Axles??

  1. #1 Finding Tongue Weight on a Trailer with Multiple Axles?? 
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    Can anyone please help me in figuring out how to find the weight placed on multiple points / fulcrums. I already know the formula for two points: (total weight) x (dist. from CG to fulcrum) / (distance between fulcrum points) = (weight on fulcrum a or fulcrum b). Now I have been trying to figure out how this changes once I have more fulcrum points. I would greatly appreciate anyone who can help!

    A real world application would be finding the weight on the trailer tongue weight and then the weight placed on axle one and axle two and so on.
    For example: I have a total weight of 3,353 lbs. The center of the axles are 130.563" from the coupler. The entire CG is 15.453" toward the coupler from the center of the axles. The axles are 33.5" apart.
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    Last edited by etheman16; November 20th, 2011 at 10:19 PM. Reason: Attached picture
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  3. #2  
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    You just have two fulcrums instead of one. You know where they are. You just have to do a little work to calculate it.


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  4. #3  
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    First you write the force balance equation. Since the trailer is not accelerating vertically, the sum of the weight on the axles plus tongue is equal to the weight of the load.
    Wt+Wa1+Wa2 = 3353 lb
    where Wt = tongue weight
    Wa1= weight on axle 1
    Wa2= weight on axle 2

    Now write an equation for the torque. You can choose any convenient point about which to calculate torque. Since there is no rotational acceleration, the torques have to add to zero. Calculating the torque about the hitch ball gives you:
    Clockwise torque = counterclockwise torque
    3353 lb* (130 9/16 - 15.453)in. =Wa1 (130 9/16-16.75) +Wa2 (130 9/16+16.75)

    Now we have 2 equations but 3 unknowns , which is a problem. I think you have to make an assumption here, like the weight is equally distributed between axle 1 and axle 2 (Wa1=Wa2).

    Just to add to the above. If the axles are rigidly attached to a rigid trailer, the distribution of weight will be highly dependent on the evenness of the road and height of the trailer hitch. For example, if the hitch is a little bit high, you would lift the front axle off the road and put all the weight on the rear axle. If a bit low, all the weight would be on the front axle.

    If the axles are suspended on identical springs, and sitting level, the springs would be compressed equally. This would justify our assumption that Wa1=Wa2.
    Last edited by Harold14370; November 21st, 2011 at 08:56 AM. Reason: Add further discussion
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  5. #4  
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    maybe think of teh wheels and hitch as negative weights?
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  6. #5  
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    Quote Originally Posted by granpa View Post
    maybe think of teh wheels and hitch as negative weights?
    The trailer hitch and axles exert a force in reaction to the weight of the trailer. If you define "down" as positive then these upward forces are negative.
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  7. #6  
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    if this is like problems with 2 fulcrums
    then it shoud be possible to consider the hitch as a fulcrum (and the other 2 points as weights)
    and then more weight will be on the first wheel

    it should also be possible to consider the first wheel to be a fulcrum and it should produce the same result.

    thirdly you can consider the second wheel to be a fulcrum and likewise it should produce the same result
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  8. #7  
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    Quote Originally Posted by granpa View Post
    if this is like problems with 2 fulcrums
    then it shoud be possible to consider the hitch as a fulcrum (and the other 2 points as weights)
    and then more weight will be on the first wheel
    What makes you think so?
    it should also be possible to consider the first wheel to be a fulcrum and it should produce the same result.

    thirdly you can consider the second wheel to be a fulcrum and likewise it should produce the same result
    Let's see your calculation.
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  9. #8  
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    Below are things that I have tried. I am thinking this might be solvable with two different formulas, one with the current formula for determining tongue weight of a single axle and then another, maybe totally different formula for determining the relationship between the two axles.

    One thing I tried (as I believe was stated above) was to take the formula for a single axle and use it by taking the total weight and dividing it between the axles and then running the formula as each axle being the center of axles instead of the center of axle being between the the axles. The problem with that is the differing tongue weights and that the total weight exceeded the actual total weight. I believe that since the axles are a part of one object that this formula theory of using the single axle formula twice for one object didn't come out right.

    The way I have it currently set up is to use the formula for the single axle trailer and use it for the tandem axle at the center of the axles. Then take the total weight and subtract the tongue weight and divide that number by the number of axles. Then take the total weight and divide it by the distance from the center of axles to the coupler and multiply this number by the distance from center of axles to each axle. After that, add the number to the to the weight opposite of the tongue weight to the axle towards the center of gravity and subtract it from the weight on the axle farthest from the CG. The problem with that is when the axles are centered on the CG, the weights don't equal each other and the total weight.

    In my mind it makes sense that if the CG is directly over the center of the axles then there will be zero tongue weight and the the total weight should be evenly distributed between the axles. As the CG is shifted more towards one axle or the other, then the weight will be differing between the axles, with more weight being applied to the axle closest to the CG and less being applied to the one farthest. With this in mind the tongue weight would begin to be positive or negative at a slower speed than if there was only one axle. I just can't figure out how this changes once another axle is in the mix. It makes sense that the tongue weight will be lighter than with a single axle. I have tested this same formula on both a finished single axle trailer (with a resulting actual tongue weight of 232 lbs. and the formula at 238 lbs.) and a tandem axle trailer, but this time the result was (actual weight of 284.5 lbs and formula at 326 lbs.) So with a difference of 41.5 lbs there should be close to equal weight placed on each axle and this is with the CG being 4.507" ahead of the front axle. As to reverse engineering this formula to come out to being even closer to the true weight, I am at a stand still. Attached are the details of the trailer that was actually tested.

    The weight on the jack was 284.5 lbs and the jack is 125.25" from the center of the axles. The total weight is 1,920 lbs.

    Trailer side.jpg
    Last edited by etheman16; November 21st, 2011 at 09:10 PM. Reason: More questioning thoughts.
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  10. #9  
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    I agree that when you factor in the springs of the shock absorbers
    you will probably wind up with the weight evenly distributed between the wheels
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  11. #10  
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    Quote Originally Posted by etheman16 View Post
    Below are things that I have tried. I am thinking this might be solvable with two different formulas, one with the current formula for determining tongue weight of a single axle and then another, maybe totally different formula for determining the relationship between the two axles.

    One thing I tried (as I believe was stated above) was to take the formula for a single axle and use it by taking the total weight and dividing it between the axles and then running the formula as each axle being the center of axles instead of the center of axle being between the the axles. The problem with that is the differing tongue weights and that the total weight exceeded the actual total weight. I believe that since the axles are a part of one object that this formula theory of using the single axle formula twice for one object didn't come out right.

    The way I have it currently set up is to use the formula for the single axle trailer and use it for the tandem axle at the center of the axles. Then take the total weight and subtract the tongue weight and divide that number by the number of axles.
    Stop right there. You're done. Everything after this is BS, because as I already explained, the weight should be evenly divided between the two axles. There are a few reasons why it might not, but if your trailer is perfectly level, and the springs are equally stiff, it will be.

    Think about it. If you have springs on each axle, and if the springs are compressed equally, each spring will be exerting the same force on its axle. If the trailer sits level, the springs will have equal compression. Otherwise the trailer wouldn't be level. There is no other way around it. Hooke's law.

    Then take the total weight and divide it by the distance from the center of axles to the coupler and multiply this number by the distance from center of axles to each axle.
    After that, add the number to the to the weight opposite of the tongue weight to the axle towards the center of gravity and subtract it from the weight on the axle farthest from the CG. The problem with that is when the axles are centered on the CG, the weights don't equal each other and the total weight.

    In my mind it makes sense that if the CG is directly over the center of the axles then there will be zero tongue weight and the the total weight should be evenly distributed between the axles. As the CG is shifted more towards one axle or the other, then the weight will be differing between the axles, with more weight being applied to the axle closest to the CG and less being applied to the one farthest. With this in mind the tongue weight would begin to be positive or negative at a slower speed than if there was only one axle. I just can't figure out how this changes once another axle is in the mix. It makes sense that the tongue weight will be lighter than with a single axle. I have tested this same formula on both a finished single axle trailer (with a resulting actual tongue weight of 232 lbs. and the formula at 238 lbs.) and a tandem axle trailer, but this time the result was (actual weight of 284.5 lbs and formula at 326 lbs.) So with a difference of 41.5 lbs there should be close to equal weight placed on each axle and this is with the CG being 4.507" ahead of the front axle. As to reverse engineering this formula to come out to being even closer to the true weight, I am at a stand still. Attached are the details of the trailer that was actually tested.

    The weight on the jack was 284.5 lbs and the jack is 125.25" from the center of the axles. The total weight is 1,920 lbs.

    Trailer side.jpg
    Last edited by Harold14370; November 22nd, 2011 at 12:23 AM.
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  12. #11  
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    Thanks! That does make sense now! I just got hung up on the idea of each axle as an independent fulcrum point and not one.


    The only thing now is that the actual tongue weight will always be lighter because the fulcrum is now spread out over 33.5" instead of on just one point. So what formula would take this into account so the 42 lb. gap in overall tongue weight can be closed? I did try taking the overall weight and subtracting the tongue weight and dividing this by the distance between the two axles: (1920 - 326) / 33.5 = 48, then 326 - 48= 278, which is 6 lbs off the actual weight and good enough for me. Is this mathematically correct though or is it just a "quick fix" that came out good once type thing?
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  13. #12  
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    Quote Originally Posted by etheman16 View Post
    Thanks! That does make sense now! I just got hung up on the idea of each axle as an independent fulcrum point and not one.


    The only thing now is that the actual tongue weight will always be lighter because the fulcrum is now spread out over 33.5" instead of on just one point. So what formula would take this into account so the 42 lb. gap in overall tongue weight can be closed? I did try taking the overall weight and subtracting the tongue weight and dividing this by the distance between the two axles: (1920 - 326) / 33.5 = 48, then 326 - 48= 278, which is 6 lbs off the actual weight and good enough for me. Is this mathematically correct though or is it just a "quick fix" that came out good once type thing?
    No, you would use the one fulcrum which is half way between the sets of wheels. In my first post on this thread, I showed how to prove that. If you find that there is a difference between the theoretical tongue weight and actual there could be several reasons for that. One is that either the road or the trailer is not perfectly level. The other is that the springs are not identical. If the front suspension is weaker than the rear, that would tend to shift the load to put more load on the rear wheels, and the tongue. This is because the effective fulcrum point is shifted toward the rear. And vice versa.
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  14. #13  
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    Can you help me in this . I did not understand fully ; How LCG had been calculated..In my case I have trailer which 3600mm long; axis centre of the two axles is at 2000mm from the front end of the trailer; axles are 800mm apart from centre axis of axle. I know the weight of the trailer is 1750Kg (it does not include axles weight) ; their is additional weight of 550 kg at a distance of 350mm from rear end of the Trailer
    Now I want to know the Tongue weight ???
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  15. #14  
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    Hi Harold14370, plz comment on above post .. Need ur help ...thx
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  16. #15  
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    Quote Originally Posted by Sunny5 View Post
    Can you help me in this . I did not understand fully ; How LCG had been calculated..In my case I have trailer which 3600mm long; axis centre of the two axles is at 2000mm from the front end of the trailer; axles are 800mm apart from centre axis of axle. I know the weight of the trailer is 1750Kg (it does not include axles weight) ; their is additional weight of 550 kg at a distance of 350mm from rear end of the Trailer
    Now I want to know the Tongue weight ???
    As we did in post 3 above,
    Wt+Wa1+Wa2 = 1750+550 kg
    where Wt = tongue weight
    Wa1= weight on axle 1
    Wa2= weight on axle 2
    We assume that the weight is evenly distributed between the two axles Wa1=Wa2 = Wa
    Wt=2300-2*Wa.
    Now write an equation for the torque. You can choose any convenient point about which to calculate torque. Since there is no rotational acceleration, the torques have to add to zero. Calculating the torque about the hitch ball (Assuming the hitch ball is at the front of the 3600 mm trailer, and weight of the trailer is evenly distributed over the 3600 mm gives you:
    The sum of the torques is zero:
    Torque due to trailer weight = 1750*1800=3150000
    Torque due to 550 kg = 550*(3600-350) = 1787500
    Torque due to rear axle = -(Wa*2400)
    Torque due to front axle = -(Wa*1600)
    3150000+1787500-Wa*4000=0
    Wa=1234.375
    Wt=2300-2*1234.375=-168.75
    The trailer is tending to lift up the tongue.

    Does that seem right? Check the math.

    ETA: I messed it up last night. I think it's right now.
    Last edited by Harold14370; January 28th, 2014 at 05:26 AM. Reason: Corrected error
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  17. #16  
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    Thanks Harold14370!
    I was confused with the position of Center of Gravity bt u made it simple. Thanks a tonne
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  18. #17  
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    Quote Originally Posted by Sunny5 View Post
    Thanks Harold14370!
    I was confused with the position of Center of Gravity bt u made it simple. Thanks a tonne
    You're welcome. We could have found the center of gravity, but this way was a little easier for me.
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