double
payload weight
Postmaster
Joined: Apr 2006
Posts: 3,524
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From: CA Bay Area
That obviously ignoring air drag and assuming the headlights have bulletproof lenses.
Postmaster
Joined: Apr 2006
Posts: 3,524
Likes: 2
From: CA Bay Area
In the Holiday spirit I decided to give another try in education on this forum. I am no teacher, but I can make drawings and even with all the scanning and downloading they don't come perfect, here they are.
. First drawing shows the truck sagging, the second leveled (lifted) by airbags.
This sample shows low and heavy load. Leveling the truck (pumping the air bags) moved the load (and the bed with cabin) pivoting it around front axle.
Y1=Y2 what means X1=X2
So by airing the bags the center of gravity did not move horizontally. Driver might be more comfortable with leveled truck, but higher center of gravity is not good for stabilization.
Postmaster
Joined: Apr 2006
Posts: 3,524
Likes: 2
From: CA Bay Area
This is how it works with high load. (that is what bpounds was talking about, I think)
Y2 is bigger than Y1, so X2 is smaller than X1. The center of gravity moved forward.
I did some quick calculation. On 10,000 lb GVW with 10 feet wheelbase, center of gravity 6 feet from front axle and 1 foot above the pivoting point the center of gravity moves 0.0173' forward after 2" lift, what equals adding 18 lb to front axle
Postmaster
Joined: Apr 2006
Posts: 3,524
Likes: 2
From: CA Bay Area
This sample shows truck with heavy load under the frame, like a truck with hidden towing boom we have in this topic.
Y2 is smaller than Y1, so X2 is bigger than X1, meaning airing the bags move the center of gravity to the rear, making the front axle having less weigh on it.
Let me highlight it again, that 10 or even 20 lb differences on axles are not going to be noticeable on handling and the main gain of air bags is driver comfort.
That still assuming that originally the suspension was not overloaded and wasn't bottoming up.
Post Fiend
Joined: Oct 2002
Posts: 11,314
Likes: 4
Assuming we are talking about a standard fixed wing aircraft with fixed engines that only produce forward thrust, it could take off. The aircraft's airspeed is determined by the amount of thrust produced by the engines and the amount of drag produced by the airframe. The airframe will require a specific airspeed to generate enough lift to lift a given weight.
There are essentially three different speeds in play here, the aircraft's ground speed (speed in relation to planet earth), the runway speed (identical to the aircraft's ground speed, but in the opposite direction on the conveyor runway), and the airspeed. Assuming there is no wind, ground speed and airspeed can be equal. Assume aircraft X has to have an airspeed of 150 mph to lift off on a normal runway, and it takes Y percent throttle to achieve this speed, this gives it a ground speed of 150 mph. Being that it is on a normal runway, the ground speed is equal to the runway speed which is equal to the airspeed. The aircraft's wheel rotate forward at 150 mph.
An aircraft applies its motive force to the air, not the ground. So if it is on the conveyor runway, and the pilots give it Y throttle, ceteris paribus, the aircraft will still move forward with 150 mph airspeed and 150 mph ground speed. The runway speed will be 300 mph as the runway is moving in the opposite direction at the same speed. The aircraft's wheels rotate forward at 300 mph. Asside from increased rolling resistance from the wheels due to their higher rpm, nothing will change.
Now if this were a car, since a car applies its motive force to the ground, it would simply sit still as long as the conveyor speed was equal and opposite to the cars wheel speed.
This is a long post for a simple question, but it's late and my thoughts consist of endless ramblings. Enjoy, lol.
There are essentially three different speeds in play here, the aircraft's ground speed (speed in relation to planet earth), the runway speed (identical to the aircraft's ground speed, but in the opposite direction on the conveyor runway), and the airspeed. Assuming there is no wind, ground speed and airspeed can be equal. Assume aircraft X has to have an airspeed of 150 mph to lift off on a normal runway, and it takes Y percent throttle to achieve this speed, this gives it a ground speed of 150 mph. Being that it is on a normal runway, the ground speed is equal to the runway speed which is equal to the airspeed. The aircraft's wheel rotate forward at 150 mph.
An aircraft applies its motive force to the air, not the ground. So if it is on the conveyor runway, and the pilots give it Y throttle, ceteris paribus, the aircraft will still move forward with 150 mph airspeed and 150 mph ground speed. The runway speed will be 300 mph as the runway is moving in the opposite direction at the same speed. The aircraft's wheels rotate forward at 300 mph. Asside from increased rolling resistance from the wheels due to their higher rpm, nothing will change.
Now if this were a car, since a car applies its motive force to the ground, it would simply sit still as long as the conveyor speed was equal and opposite to the cars wheel speed.
This is a long post for a simple question, but it's late and my thoughts consist of endless ramblings. Enjoy, lol.
we have a winner...
the name of the device "AIRplane" tells you the answer...
it's not a groundplane...
it's engines act on AIR - not ground
it's thrust is the force applied to the air around the plane -
the wheels/ground/conveyor factoids are irrelevant
when I first heard of this years ago, there wasn't much out on the net about it... now it's all over the place - do a google search
but to the OP, it is what it is and you have to make it work however you can...
that's what I am doing !
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Senior User
Joined: Aug 2003
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From: SW ND
Kajtek, why did you put a defined pivot point in those diagrams? Should have have put into the equation of front spring rates. Doesn't that frame transer weight on way or the other?
I can't follow your logic, because it doesn't jive with my hands on experience, both in my pickup and in a heavy haul truck.
I can't follow your logic, because it doesn't jive with my hands on experience, both in my pickup and in a heavy haul truck.
Postmaster
Joined: Apr 2006
Posts: 3,524
Likes: 2
From: CA Bay Area
Kajtek, why did you put a defined pivot point in those diagrams? Should have have put into the equation of front spring rates. Doesn't that frame transer weight on way or the other?
I can't follow your logic, because it doesn't jive with my hands on experience, both in my pickup and in a heavy haul truck.
I can't follow your logic, because it doesn't jive with my hands on experience, both in my pickup and in a heavy haul truck.
Can't quite understand what is your experience, but laws of physics and geometry are pretty stiff laws. There is no way to bend them. No matter how much salesman would try.
Senior User
Joined: Aug 2003
Posts: 457
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From: SW ND
What do you consider a minimal change in suspension sag?
What weight will make a minimal change in suspension sag?
What do you consider a large amount of sag?
What weight will make a large amount of sag?
What weight will make a minimal change in suspension sag?
What do you consider a large amount of sag?
What weight will make a large amount of sag?
Postmaster
Joined: Apr 2006
Posts: 3,524
Likes: 2
From: CA Bay Area
What do you consider a minimal change in suspension sag?one done by 18lb
What weight will make a minimal change in suspension sag? see above
What do you consider a large amount of sag? in the sample above I predicted 2" of lift
What weight will make a large amount of sag? that doesn't matter for the law of physics. The drawing above will apply to lawnmower and to semitruck as well. Gave one calculation that would apply to SD, but that only to put a number to the dispute.
What weight will make a minimal change in suspension sag? see above
What do you consider a large amount of sag? in the sample above I predicted 2" of lift
What weight will make a large amount of sag? that doesn't matter for the law of physics. The drawing above will apply to lawnmower and to semitruck as well. Gave one calculation that would apply to SD, but that only to put a number to the dispute.
Elder User
Joined: Jun 2011
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If the center point under your weight (the load you are hauling into the bed of your truck) is over your rear axel, then at that position is where the fulcrum is located. NOT over the front axel. Your pivot point should be the rear axel.
Postmaster
Joined: Apr 2006
Posts: 3,524
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From: CA Bay Area
And how exactly you measure weight distribution with your tape measure?