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Engineering math question for you

6K views 83 replies 35 participants last post by  murf 
#1 ·
Does anyone happen to know the formula that gives you the amount a weight a vehicle loses as it goes faster down the road? I am not talking about orbiting the earth or escape velocity etc. Just a simple formula where you enter the vehicle weight and speed and end up with an answer. Smarta$$ answers are not appreciated. I am asking a serious question please for a young man doing a science experiment.
 
#2 ·
I'm pretty sure there is no generic formula. It will depend upon the shape of the vehicle under consideration. If the vehicle produces a low-pressure above the vehicle, it will produce lift and lessen the effective weight of the vehicle as speed increases. The converse is also true. Some vehicles actually increase the downforce as the speed increases.

It all depends upon the aerodynamics of the vehicle in question.

Smarter minds chime in here.

Glen
 
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#3 ·
First, @Tom Finch would be the one to ask. Being as he used to be involved in aircraft design and has some patents related ... there is a good reason that his Tailwind trailers kind of look like an aircraft wing.

I am guessing that there is not a constant formula for that. The aerodynamic design (or lack there of) of the vehicle has a major affect that.
 
#4 ·
I wasn't considering aerodynamics in the question. I read where mass actually increases with speed. The science experiment for the young boy is supposed to determine if the weight of a car going down the road at say 100MPH weighs less due to centrifugal force around the earth. All other considerations like aerodynamics would be getting too complicated for him. All help is apppreciated.
 
#11 · (Edited)
I'm an electrical engineer, not a physicist, but that never stopped me before.......

This is essentially a centripetal force question. To give meaningful results in the experiment, the answer pretty much has to ignore the effects of the surrounding air. In other words, in a vacuum, since the air can cause either an effective downwards force or upwards (lift), depending on the design of the vehicle.

A vehicle effectively becomes weightless on Earth at Mach 33, or about 25,000 miles per hour. In other words, it completely overcomes the effect of gravity at that speed, also known as escape velocity. I have not looked it up, but the formula for calculating escape velocity is probably where you are going to want to start with this.

In this particular scenario, I do not think a vehicle traveling at a relatively slow speed of 100 mph is even affected by this, because the mass of the earth and its radius are far too great for centrifugal force to even begin to have an effect on the weight of the bike.

Formulas for orbit velocity are readily available on any physics website, but that doesn't help, because the mass or weight of the vehicle does not factor in. It might require some real digging to come up with a formula for this particular problem.
 
#5 ·
The only change to weight would be related to some kind of aerodynamic effect causing either lift or down force on the vehicle.

Outside of that, the constant here is gravity and weight is just the affect of gravity on the mass. If gravity is unchanged and the mass of the vehicle is unchanged, then the weight is unchanged.
 
#6 ·
I suggested an experiment for the young man. I proposed that he put a motorcycle (Not my wing!) in the back of a truck with bathroom scales beneath each wheel. We could go down a back road at a given speed and see if the weight changed on the scales while moving. I suspect it will not, but thought I would ask just in case I could be wrong. I haven't figured out the details yet on how to stabilize the bike other than simply using the side stand. Thinking of something like a small dirt bike. Maybe it will change but be so little to actually read anything. Just trying to help the kid.
 
#7 · (Edited)
Oh, I see his point now. Like a space craft in stable, near-earth orbit. If it accelerates enough, it will escape earth's gravity and go off into outer space.

Assume a car has zero aero lift or down force. How fast would it have to go to "lift off" like the space ship?

On my '70 Dodge Dart the answer was 140mph, but I think that was aero-lift! Lucky to have survived!

Send Tom a PM.
 
#8 ·
As we sit here looking at our device screens we are moving at tremendous speed. The earth makes a complete revolution on it axis in 24 hours. The earth every year makes a complete orbital path around the sun. It is believed that our entire galaxy is moving through space at some tremendously fast speed. But we don't "feel" the speed at which we are presently moving, although it is certainly recognizable when we look to the sky. I suspect there is no detectable change in mass relative to speed, especially at speeds as slow as 100 mph, if indeed there is any change of mass at any speed.

I've already typed more than I know:nerd:
 
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#10 ·
This reminds me of an advanced math problem in high school. If you could fire a rifle bullet and drop a piece of chalk of the same weight at the same time, assuming the bullet is traveling in a straight line, they would both hit the ground at the same time.
 
#12 · (Edited)
Some stuff is cross-threaded here.

Stop thinking about weight. Force = mass x acceleration. Weight is basically force of your mass under the acceleration of gravity. Think mass. Unless you're getting into stuff like consumption of fuel over a small term, mass is going to be constant.

Aerodynamics could add acceleration but you seem happy to assume those away (good idea). And unless you're getting up around the speed of light, skip relativity.

So it boils down to how what you're doing is going to change the force of gravity. I'm pretty sure it's not going to. I'm not aware of a simple equation for that. And I'm not sure what you're looking for.

BTW, there isn't really a centrifugal force in this case since we're not talking about, for example, a string that keeps something that's spinning. There is a centripetal (away) force caused by the rotational velocity, but the force of gravity exceeds that. If you juice up on that with your booster rockets or your Goldwing tuned up a bit, you can get your centripetal up to the point where it exceeds gravity and then you can break away. But don't try that at home.

There is something called the Coriolis Effect that has to do with deflection of paths as things move across the earth and get faster or slower relative to their neighbors that are spinning in orbits that are slightly further or closer to the equator. That causes hurricanes and toilets to spin the ways they do and artillery folks to apply a correction to their firing data. Not mass related.

Good luck.

P.S. didn't see Sparky's post until after I hit reply. Perhaps that angle is the trick -- what percentage of escape velocity have I achieved as a lightening effect, even though the OP says not interested in escape velocity? Note that you need to be driving in the right direction, go the wrong way and you'll get heavier. It also gets semantic on what weight is. The sum of all forces on an object can be zero but it still has weight being countered by an equal up force (the centripetal force of orbital acceleration? the fan at the sky dive simulator?).

P.P.S. D'OH. What a dummy. I was up on my soap box and switched those centripetal and centrifugals around. They're reversed above, left them wrong and just edited to highlight rather than sweep over my mistake. Now which one is a stalagmite?
 
#16 ·
D'oh! Yes, I mixed up centripetal and centrifugal before. Dummy.

Spinning wheels create gyroscopic stability. Torque, procession, and spin form a right handed coordinate system - that describes which way they react to a force that tries to deflect the spinning wheel. Helps us stay upright, and accounts for press to lean. Googling those terms will get you TMI.

Check this out. Perhaps this may help.

http://www.phy.olemiss.edu/HEP/quarknet/mass.html

:doorag:
Better be a fast truck!
 
#18 · (Edited)
Y'all are doing pretty well on this. The original question does not allow for the effects of aerodynamics which will make the car either heavier, lighter or no appreciable change depending on the vehicle. My Goldwing gets very light and floaty at 100+, when I raced corvettes the vehicle got heavier - both effects due to aerodynamics.


As indicated - there is a difference between centrifugal and centripetal force. Now as for the vehicle in question, if aerodynamics is removed from the equation - then the question remains - "Does the vehicle weigh less going 100 mph?"


As discussed earlier by 2wheelnut - if mass and gravity remains constant - then the weight remains constant. However, in this case the vehicle is traveling at speed. Now if you'll remember Newton - an object in motion remains in motion unless acted upon by an outside force. The corollary is an object will travel in a straight line unless acted upon by an outside force. So the vehicle will try to fly into space, just like a bullet fired from a gun will try to fly into space. Again - that nasty force called gravity comes into play (the outside force described above).


Now - you get to use some math to determine how much angular force is required to overcome the outward inertia imparted by the 100 mph vehicle (speed is just an example).




edit: sorry about the formatting inclusions - copy and paste :-(


Formula[edit]

The magnitude of the centripetal force on an object of mass m moving at tangential speed v along a path with radius of curvature r is:[6]
F = m a c = m v 2 r {\displaystyle F=ma_{c}={\frac {mv^{2}}{r}}}
where a c {\displaystyle a_{c}}
is the centripetal acceleration. The direction of the force is toward the center of the circle in which the object is moving, or the osculating circle (the circle that best fits the local path of the object, if the path is not circular).[7] The speed in the formula is squared, so twice the speed needs four times the force. The inverse relationship with the radius of curvature shows that half the radial distance requires twice the force. This force is also sometimes written in terms of the angular velocity ω of the object about the center of the circle, related to the tangential velocity by the formula
v = ω r {\displaystyle v=\omega r}
so that
F = m r ω 2 . {\displaystyle F=mr\omega ^{2}\,.}
Expressed using the orbital period T for one revolution of the circle,
ω = 2 π T {\displaystyle \omega ={\frac {2\pi }{T}}\,}
the equation becomes
F = m r ( 2 π T ) 2 . {\displaystyle F=mr\left({\frac {2\pi }{T}}\right)^{2}.}
[8]In particle accelerators, velocity can be very high (close to the speed of light in vacuum) so the same rest mass now exerts greater inertia (relativistic mass) thereby requiring greater force for the same centripetal acceleration, so the equation becomes:
F = γ m v 2 r {\displaystyle F={\frac {\gamma mv^{2}}{r}}}
where
γ = 1 1 − v 2 / c 2 {\displaystyle \gamma ={\frac {1}{\sqrt {1-v^{2}/c^{2}}}}}
is called the Lorentz factor.
More intuitively:
F = γ m v ω {\displaystyle F=\gamma mv\omega }
which is the rate of change of relativistic momentum (γ m v {\displaystyle \gamma mv}
)


Additional info can be found - https://en.wikipedia.org/wiki/Centripetal_force


Reminder - the mass of the vehicle remains constant, the weight is reduced at speed due to the angular effect of motion tangential to the force of gravity.
.
 
#21 ·
Is this along the same line of thought regarding the tire on our motorcycle and it's ability to stay in contact with the ground? Relatively speaking our tires get less grip the faster we go, then add in turning or braking, or both, until we exceed it's ability to grip the surface? granted there are some changes to this compared to that, but it's sounds similar to my brain. Good job SSNOB.
 
#20 ·
My $.02 is they have scales that can weigh a semi on the move to check if over weight. They may not be going 100 mph but at 70 mph they check for over weight trucks. That's about as mathematical as I can get.
 
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#28 ·
This discussion reminds me of the question "why does a house fly or bee buzzing around the inside my truck when I'm travelling 70 mph down the highway not get splattered on the inside side of my windshield?"
 
#44 ·
I could still use one for the basic stuff, anything more than that I'd have to pull out the instruction sheet. :wink2:

I think this is, as suggested by ssncob, a question of angular momentum. For example we all know that the astronauts experience weightlessness when orbiting. Well, it isn't really weightlessness, it is that the rate of fall is offset by the angular velocity of the space craft. For every foot it falls, the space ship travels fast enough perpendicular to the plain of gravity to offset the fall. If the spacecraft velocity were zero relative to the center of the earth, the spaceship would immediately begin falling and the astronauts would experience weight once the spacecraft entered the atmosphere.

So, the original question posed is legitimate if asking how much acceleration of gravity is offset by the angular momentum of a Goldwing. The piece of information not provided is which direction is the Goldwing traveling? You may have heard that space launch ports are as close to the equator as possible. This is because the earth, rotating eastward, has the greatest velocity at the equator. So how much benefit does the almost 1,000 miles/h rotation have on launch requirements? It reduces the effective gravity by about 0.346%. Another way of saying this is people at the equator experience 99.654 percent of the gravitational pull as those at the polls (this does not factor in earths bulge, gravitational anomalies or changes in elevation). Not much for an additional 1,000 miles/h. Should your Goldwing be headed east, it will add to the angular momentum and reduce the perceived weight, but head west and it will subtract from the earths angular momentum actually making your Goldwing heaver. Variations in direction in velocity to North or South, would need to be calculated relative to absolute direction.

Fun thought puzzle.

I love your description. I totally forgot to take into account the east-west rotational impact - was looking at it strictly from an angular momentum perspective. :thumbup:




The weight of the vehicle does not change no matter what speed it is going. Only the forces acting on it may change, either positive force or negative force, but the weight remains the same. Weight is a factor of gravity acting on mass. The mass of the vehicle does not change with speed.
I agree.... weight is the force of gravity acting on the mass.... all other forces resulting from aerodynamics, bumps and hills do generate forces that act on the mass but these are not forces caused by gravity. As a result these other forces cannot change the weight. The weight can ONLY be changed by changes in gravity or changes in mass.

Gravity .... here on earth is constant for all practical purposes.
...
So I say the answer is zero. The weight will not change as the motorcycle speeds down the road and is a constant. The weight is not a function of speed.

Y'all are right when you say mass does not change (again not counting for fuel consumption, parts falling off, etc :)) However you are incorrect when saying the weight does not change, or that gravity is constant here on earth. As mentioned earlier in this thread - (force = mass x acceleration). Now mass does not change, but acceleration due to gravity does. When calibrating force devices such as load cells and torque cells - if using a deadweight system then the local gravity does need to be taken into account. The weights are calibrated and assigned a value corrected to standard gravity (9.80665 m/s2 ). When using the weights at different locations, the local gravity constant must be accounted for. My testing lab was 0.03 % lighter than standard gravity, therefore 100 lbs of deadweight applied to a force cell was actually 99.97 lbs of force. May not seem like much of a difference - but it matters. Here's a link for gravity contour plotting for the U.S. - https://mrdata.usgs.gov/geophysics/gravity.html

I suspect that the weight will change dependent on the velocity since the earth is not flat and there will be a tendency for a moving object to travel in a straight line. As the velocity increases there will be more of a tendency for an object to straighten its trajectory and eventually it will travel in a curvature less than that of the earth's surface curvature and therefore it would begin to rise from the earth's surface. At a point a little before it actually begins to rise it will be in equilibrium to gravity and, therefore, weightless.
Exactly - that's why roller coasters give you that weightless feeling, and the Vomit Comet trains astronauts.
 
#32 ·
I think this is, as suggested by ssncob, a question of angular momentum. For example we all know that the astronauts experience weightlessness when orbiting. Well, it isn't really weightlessness, it is that the rate of fall is offset by the angular velocity of the space craft. For every foot it falls, the space ship travels fast enough perpendicular to the plain of gravity to offset the fall. If the spacecraft velocity were zero relative to the center of the earth, the spaceship would immediately begin falling and the astronauts would experience weight once the spacecraft entered the atmosphere.

So, the original question posed is legitimate if asking how much acceleration of gravity is offset by the angular momentum of a Goldwing. The piece of information not provided is which direction is the Goldwing traveling? You may have heard that space launch ports are as close to the equator as possible. This is because the earth, rotating eastward, has the greatest velocity at the equator. So how much benefit does the almost 1,000 miles/h rotation have on launch requirements? It reduces the effective gravity by about 0.346%. Another way of saying this is people at the equator experience 99.654 percent of the gravitational pull as those at the polls (this does not factor in earths bulge, gravitational anomalies or changes in elevation). Not much for an additional 1,000 miles/h. Should your Goldwing be headed east, it will add to the angular momentum and reduce the perceived weight, but head west and it will subtract from the earths angular momentum actually making your Goldwing heaver. Variations in direction in velocity to North or South, would need to be calculated relative to absolute direction.

Fun thought puzzle.
 
#36 · (Edited)
I agree.... weight is the force of gravity acting on the mass.... all other forces resulting from aerodynamics, bumps and hills do generate forces that act on the mass but these are not forces caused by gravity. As a result these other forces cannot change the weight. The weight can ONLY be changed by changes in gravity or changes in mass.

Gravity .... here on earth is constant for all practical purposes.

Mass.... the only thing that can change the mass are fuel consumption, parts fall off, birds poop on you and bugs splat and stick to the windshield. ... stuff like that.

The original poster did not offer any specifics on fuel consumption. Just mentioned the motorcycle speeds down the road.


So I say the answer is zero. The weight will not change as the motorcycle speeds down the road and is a constant. The weight is not a function of speed.
 
#37 · (Edited)
I suspect that the weight will change dependent on the velocity since the earth is not flat and there will be a tendency for a moving object to travel in a straight line. As the velocity increases there will be more of a tendency for an object to straighten its trajectory and eventually it will travel in a curvature less than that of the earth's surface curvature and therefore it would begin to rise from the earth's surface. At a point a little before it actually begins to rise it will be in equilibrium to gravity and, therefore, weightless.
 
#38 ·
An airplane flying overhead has the same weight as one parked on the tarmac. When it's flying, aerodynamic forces (lift) are holding it up. The wings are supporting it,

When the airplane is on the tarmac.... the ground is supporting it. In both cases the plane weighs the same.

It's mass is the same in both locations and gravity is the same in both locations. Therefore it's weight is the same.

Weight has nothing to do with if something is on the ground or not.
 
#40 ·
So if an aircraft takes off with 100 eagles standing on the floor, and assuming the eagles way 10 pounds each, once airborne if the eagles fly inside the aircraft, no longer touching the aircraft, does the aircraft weigh 1000 pounds less?
 
#42 ·
So if an aircraft takes off with 100 eagles standing on the floor, and assuming the eagles way 10 pounds each, once airborne if the eagles fly inside the aircraft, no longer touching the aircraft, does the aircraft weigh 1000 pounds less?
No, the total weight contained inside the airplane is still the same......in order for them to fly they are pressing their wings against the air which is pressing against the plane.

If they were to leave the plane that is a different story.
 
#41 ·
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