Anywhere in the UK you can take your Zed to see what it will do?

[john-e89]

Yes plus full tank of fuel and spare wheel

Perfect Possibly the only time the driver could actually be sure they were adding to the equation when it comes to going faster in a Zed. :wink:

Mass is irrelevant when considering acceleration due to gravity:

v^2 = u^2 + 2as

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

Beachy Head is 162m in height. Assuming that u^2 is zero (since you aren’t falling until you leave the cliff at which point you start to accelerate downwards) and ignoring drag, your velocity when you hit the beach is:

v = sq root (0 + (2 x 9.8 x 162)) = 56.3m/s = 202.9kph = 126mph

I suspect a car falling wheels down wouldn’t come close to 126mph due to drag, whilst a car falling nose first would probably exceed that if the cliffs were higher. Of course, given that most cars experience lift at high speed, if your horizontal velocity is great enough when you leave the cliff and the car remains stably perpendicular to the ground, it might even glide, landing gently and gracefully in the sea

Never mind all that cobblers.....dead is dead..... :wink:

 

[buzyg]

Yes plus full tank of fuel and spare wheel

Perfect Possibly the only time the driver could actually be sure they were adding to the equation when it comes to going faster in a Zed. :wink:

Mass is irrelevant when considering acceleration due to gravity:

v^2 = u^2 + 2as

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

Beachy Head is 162m in height. Assuming that u^2 is zero (since you aren’t falling until you leave the cliff at which point you start to accelerate downwards) and ignoring drag, your velocity when you hit the beach is:

v = sq root (0 + (2 x 9.8 x 162)) = 56.3m/s = 202.9kph = 126mph

I suspect a car falling wheels down wouldn’t come close to 126mph due to drag, whilst a car falling nose first would probably exceed that if the cliffs were higher. Of course, given that most cars experience lift at high speed, if your horizontal velocity is great enough when you leave the cliff and the car remains stably perpendicular to the ground, it might even glide, landing gently and gracefully in the sea

Not quite sure why you quote a statement for acceleration in a perfect vacuum and then go on to discuss that the a Mass does indeed increase both the Acceleration and the Terminal velocity as soon as you introduce an atmosphere.

Terminal velocity is determined by force down vs the force up. Force down is Mass x Acceleration. so the extra Mass will increase the velocity downward. Simples :wink:

 

[BMWZ4MC]

Mass definitely doesn’t change the rate of acceleration due to gravity! Heavier objects have do have a greater downward force, but they also have more inertia. As such they require more force to accelerate them. Galileo proved that over four hundred years ago.
As for terminal velocity, that’s determined by drag, buoyancy and mass. In this instance, drag would be by far the more important determinant. Besides, Beachy Head isn’t high enough to reach terminal velocity, so mass is irrelevant.

 

[buzyg]

Mass definitely doesn’t change the rate of acceleration due to gravity! Heavier objects have do have a greater downward force, but they also have more inertia. As such they require more force to accelerate them. Galileo proved that over four hundred years ago.
As for terminal velocity, that’s determined by drag, buoyancy and mass. In this instance, drag would be by far the more important determinant. Besides, Beachy Head isn’t high enough to reach terminal velocity, so mass is irrelevant.

Mass is not irrelaven. It has an effect on both the Acceleration and the Velocity, any where there is an atmosphere. To try and say otherwise is a bit daft. :wink:

 

[BMWZ4MC]

Hmmm...Galileo or Buzyg? Tough choice but your reasoned argument has won me over, I salute you sir!

 

[john-e89]

Perfect Possibly the only time the driver could actually be sure they were adding to the equation when it comes to going faster in a Zed. :wink:

Mass is irrelevant when considering acceleration due to gravity:

v^2 = u^2 + 2as

Where v = final velocity, u = initial velocity, a = acceleration, s = distance

Beachy Head is 162m in height. Assuming that u^2 is zero (since you aren’t falling until you leave the cliff at which point you start to accelerate downwards) and ignoring drag, your velocity when you hit the beach is:

v = sq root (0 + (2 x 9.8 x 162)) = 56.3m/s = 202.9kph = 126mph

I suspect a car falling wheels down wouldn’t come close to 126mph due to drag, whilst a car falling nose first would probably exceed that if the cliffs were higher. Of course, given that most cars experience lift at high speed, if your horizontal velocity is great enough when you leave the cliff and the car remains stably perpendicular to the ground, it might even glide, landing gently and gracefully in the sea

Not quite sure why you quote a statement for acceleration in a perfect vacuum and then go on to discuss that the a Mass does indeed increase both the Acceleration and the Terminal velocity as soon as you introduce an atmosphere.

Terminal velocity is determined by force down vs the force up. Force down is Mass x Acceleration. so the extra Mass will increase the velocity downward. Simples :wink:

You know what’s really really weird....I actually get what you’re saying Buzy....

Not only that it makes sense....!!

 

[M1k3yC]

You might find this interesting. It also includes a terminal velocity calculator. https://www.grc.nasa.gov/WWW/K-12/airplane/termv.html

You'd hope that NASA understands how to do this. :wink:

 

[BMWZ4MC]

That link is interesting and gives a good formula for terminal velocity, but it doesn’t address the key point being debated here:
I say that two otherwise identical bodies of different density (and so different mass) will accelerate at the same rate. Buzy says that the heavier body will accelerate faster. Buzy is from the school of Aristotle whilst I think Galileo was correct.

 

[buzyg]

You might find this interesting. It also includes a terminal velocity calculator. https://www.grc.nasa.gov/WWW/K-12/airplane/termv.html

You'd hope that NASA understands how to do this. :wink:

Just what the Doctor ordered.

 

[buzyg]

Double post. Delete

 

[Mister T]

RAF Marham used to host these kind of events. I've done quite a few 30-130 and top speed runs there.

 

[buzyg]

That link is interesting and gives a good formula for terminal velocity, but it doesn’t address the key point being debated here:
I say that two otherwise identical bodies of different density (and so different mass) will accelerate at the same rate. Buzy says that the heavier body will accelerate faster. Buzy is from the school of Aristotle whilst I think Galileo was correct.

Good to see NASA understand life on earth as well as the moon. Galileo was a genius and predicted/calculated something he could not observe. ie that two object falling in a perfect vacuum would accelerate at equal rates. Beachy head however is a place full of atmosphere. Hence there is another force opposing the acceleration. This force is related to the weight of the object. So though you get 10/10 for effort. Ultimately there is only one correct answer for a Z4 falling from Beachy head.

It's going to crash.

 

[Anonymous]

That link is interesting and gives a good formula for terminal velocity, but it doesn’t address the key point being debated here:
I say that two otherwise identical bodies of different density (and so different mass) will accelerate at the same rate. Buzy says that the heavier body will accelerate faster. Buzy is from the school of Aristotle whilst I think Galileo was correct.

Your only correct if you do the experiment in a vacuum...for reference there is the famous NASA clip of an astronaut on the moon dropping a feather and weight together..they hit the ground at the same time...aeronautical drag doesn’t exist in a vacuum...

However on earth everything suffers from drag....drag goes up as a function of square of the velocity...

So your car at 60mph had 4 times the drag at 30 mph ..

Therefore drag will be a factor in the Z4 off Beachy Head..

The NASA equation shows terminal velocity V = sqrt ( (2 * W) / (Cd * r * A)

W is weight ..so you are incorrect ...two identical,say cubes, one of steel, one of lead..ie same identical bodies but different mass ie weight ie densities will reach different terminal velocities..due to drag..

Galileo’s experiment was just that an experiment...sadly at the time he didn’t major in computational fluid dynamics..

 

[BMWZ4MC]

Again, I’m not talking about terminal velocity, where mass is relevant.
Over distances insufficient to attain terminal velocity, the rate of acceleration due to gravity of two falling objects of identical size, shape and surface but different densities will be the same in the same atmosphere.
Posts giving equations for determining terminal velocity and showing how mass influences this do nothing to support the argument that mass is of relevance at velocities less than terminal velocity.
Please provide evidence that mass determines rate of acceleration or velocity attained for a falling object at less than terminal velocity.

 

[buzyg]

Again, I’m not talking about terminal velocity, where mass is relevant.
Over distances insufficient to attain terminal velocity, the rate of acceleration of two objects in the same atmosphere and of identical size, shape and surface but different densities will be the same.
Posts giving equations for determining terminal velocity and showing how mass influences this do nothing to support the argument that mass is of relevance at velocities less than terminal velocity.
Please provide evidence that mass determines rate of acceleration or velocity attained for a falling object at less than terminal velocity.

I give up. Your completely wrong but you clearly don't understand the equations. Happy with that, there is no more I can add to convince you of the true nature of gravity and what happens when air gets in the way of and accelerating mass. It accelerates more slowly than in a vacuum by the way and the heavier it is per unit of area the faster it accelerates relative to an object of equal area but less mass. That is called weight in the NASA Equation if that helps..

 

[john-e89]

Again, I’m not talking about terminal velocity, where mass is relevant.
Over distances insufficient to attain terminal velocity, the rate of acceleration of two objects in the same atmosphere and of identical size, shape and surface but different densities will be the same.
Posts giving equations for determining terminal velocity and showing how mass influences this do nothing to support the argument that mass is of relevance at velocities less than terminal velocity.
Please provide evidence that mass determines rate of acceleration or velocity attained for a falling object at less than terminal velocity.

I give up. Your completely wrong but you clearly don't understand the equations. Happy with that, there is no more I can add to convince you of the true nature of gravity and what happens when air gets in the way of and accelerating mass. It accelerates more slowly than in a vacuum by the way and the heavier it is per unit of area the faster it accelerates relative to an object of equal area but less mass. That is called weight in the NASA Equation if that helps..

I’m reading this as two identical oil drums, one filled with lead, the other with polystyrene thrown off beachy head at the same time will accelerate at the same rate according to BMWZ4MC.....yes...?

 

[BMWZ4MC]

Buzy, I’m not being stubborn, I’d like to be educated so please explain!
Dropped not thrown off Beachy Head, John. To avoid ambiguity, let’s use a 10m cliff instead so there is no risk of reaching terminal velocity. I’m saying that the rate of acceleration and velocity attained over that 10m will be the same for both oil drums.
If you drop them from a ‘plane at 1000m then the drum of lead will continue to accelerate after the polystyrene-filled drum has reached a constant velocity. Subsequently, the heavier drum will also reach a constant velocity which will be greater than that of the lighter drum.

 

[john-e89]

Buzy, I’m not being stubborn, I’d like to be educated so please explain!
Dropped not thrown off Beachy Head, John. To avoid ambiguity, let’s use a 10m cliff instead so there is no risk of reaching terminal velocity. I’m saying that the rate of acceleration and velocity attained over that 10m will be the same for both oil drums.
If you drop them from a ‘plane at 1000m then the drum of lead will continue to accelerate after the polystyrene-filled drum has reached a constant velocity. Subsequently, the heavier drum will also reach a constant velocity which will be greater than that of the lighter drum.

See the pics, I’ve just dropped these candles from the same height several times, two taped together VS just the one, they both hit the floor at exactly the same time. One is twice as heavy as the other obvs.

Richard is right.

 

[MrPT]

Oxford Street is pretty empty at the moment.

 

[john-e89]