tabletop tripod for Nikon D80 w/ 18-135mm f/3.5-5.6 lens?? READ DETAILS?

Written by Ev in Saturday, May 12, 2007 under Cameras
Comments: 1

Q: i want a tripod that will support my camera with the heavier 18-135mm lens. basically, this is a scenario i see. while on vacation, say at a cafe, and i want the camera to be eye level with me while sitting. therefore, i need some mini/tabletop/travel tripod that can change heigths. plesae give specific models, thx.

A: The only way to find this out is to take your rig down to your local camera store and stick it on a bunch of tripods.

I have a D50 with 18-70 and none of the lightweight tabletops would hold it up. I suspect your setup will have an even worse problem.

For your application Nikon's 50mm f1.8 might be a better lens. It's tiny -- better balance on a pod -- and lets in a ton more light than the zoom so you'll get faster shutter speeds and less blur, and...

Comments: 1

mygif
Stanley W Said in Saturday, May 12, 2007

There are small tabletop tripods that have no make or model because it is very cheap. They have flexible legs. These are you best best if you do not want to splurge for Manfrotto or Bogen.

It will be tough to balance such a heavy lens but it is possible and better than nothing.

Do not bother with fixed legs type. They will not be able to support the weight.

how do you find initial horizontal velocity?

Written by angry_numbat | Sunday, May 04, 2008

I have an assignment that asks this question and i have no idea how to do it. The question is:
A ball travels across a tabletop with constant velocity, then drops a distance of 1225mm to the floor. It lands 2.45m from a point on the floor directly under the edge of the table. Calculate the balls original velocity across the table.


A: Calculate for the time the ball will reach the floor:

dy = Viy(t) + (1/2)(g)t^2

Consider downward as positve.

1.225 m = 0(t) + (1/2)(9.8 m/s^2)t^2
t = sqrt[1.225m/4.9m/s^2]
t = 0.5 sec

Calculating for Original Velocity across the table:
dx = Vx(t)
2.45 m = Vx(0.5sec)
Vx = 2.45 m/0.5 sec
Vx = 4.9 m/sec, original velocity across the table.

ANSWER: 4.9 m/sec

...

Can you solve the hardest physics quesiton in the world?

Written by Bob Loblaw | Monday, January 29, 2007

A glass ornament of mass 575g sitting on a table is subjected to a resonant frequency of 440hz. The ornament breaks into three pieces that travel horizontally across the frictionless tabletop. Fragment A has a mass of 168g and fragment B has a mass of 212g. What is the magnitude of the momentum of the third piece of glass fragment c?


A: Yes the 440hz frequency is irrelevant and yes we can't give a numeric answer without knowing the velocity (speed and direction) of the other two fragments. But we can surely give an answer that includes these unknowns as variables.

We choose our axes so that f1 (fragment 1) is moving along the x-axis, and f2 (fragment 2) is moving at an angle 'theta' to the axis.

The magnitude of the momentum is just the length of the side that completes the triangle...

whatcha know about Projectile Motion? Can you answer this one?

Written by arliss | Monday, May 15, 2006

A ball rolls off a tabletop with an inital velocity of 3 m/s, if the tabletop is 1.25m above the floor, A. how long does it take for the ball to hit the floor? B. how far does the ball travel horizontally?

please show your work....

good luck.


A: That's easy
The key to projectile motion is resolving motion horizontally and vertically(or downwards).The vertical motion is accelerated motion,acceleration being g while horizontal motion is not accelerated and is at constant speed.
Suppose at time t=0 the ball leaves the table.
Vh(0)=3m/s Horizontal Velocity
Vd(0)=0m/s Downwards velocity
Now time taken to hit floor is same as time taken by downwards velocity component to cover 1.25...

Cats + Dogs in the Sand

Source: Completely Coastal | Beach + Nautical Decorating, Art, Craft Ideas, Homes + more.

Hope you have a laugh...

I can has Cheezburger

I has a Hot Dog
Sharing with: Outdoor Wednesday

Related Post: Beach Lovin' Dogs

I haven't found an official Cat Beach yet, but there is the Dog's Beach and Jen tells the stories...

... and today's post on my new blog A Beach Lovers Place: Watch your Step!