Fulcrum problem (I think)


by ICLKennyg
Tags: fulcrum
ICLKennyg
ICLKennyg is offline
#1
Aug12-08, 09:08 PM
P: 2
I have a problem and it's been a while since I took a physics class.

So camera tripods are rated in weight... but this is more a function of torque than weight. The idea is the head can only hold so much force level. Obviously as the lens weight and length increases the force on the camera and tripod head gets worse. I am looking for a formula that will tell me what a lens will give me in a torque force on a tripod head. Sorry google doesn't seem to have much help on this.

For simplicity's sake we can assume that the camera has no thickness.

If a camera weighs 500g and a lens weighs 1kg distributed to a balancing point of 4cm away from the camera and 6cm away from the end (10cm long lens) how much torque would this generate on the tripod head. If it really matters the camera in question is 5cm wide with the mount roughly in the middle. Also in case it's important the distance from the tripod mount (bottom) to the center of the lens mount is 3cm.

A quick diagram of what I am talking about.
http://kj.stillabower.net/forum/imgs/camera.jpg
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isly ilwott
isly ilwott is offline
#2
Aug13-08, 09:16 AM
P: 72
Quote Quote by ICLKennyg View Post
I have a problem and it's been a while since I took a physics class.

So camera tripods are rated in weight... but this is more a function of torque than weight. The idea is the head can only hold so much force level. Obviously as the lens weight and length increases the force on the camera and tripod head gets worse. I am looking for a formula that will tell me what a lens will give me in a torque force on a tripod head. Sorry google doesn't seem to have much help on this.

For simplicity's sake we can assume that the camera has no thickness.

If a camera weighs 500g and a lens weighs 1kg distributed to a balancing point of 4cm away from the camera and 6cm away from the end (10cm long lens) how much torque would this generate on the tripod head. If it really matters the camera in question is 5cm wide with the mount roughly in the middle. Also in case it's important the distance from the tripod mount (bottom) to the center of the lens mount is 3cm.

A quick diagram of what I am talking about.
http://kj.stillabower.net/forum/imgs/camera.jpg
The link wouldn't work for me.

The answer appears to be quite simple to figure though. If the cg (center of gravity) of the camera is exactly over the vertical shaft of the tripod, it will present 0 gram.centimeters torque to the head. If it is not, you must add (or subtract) the torque presented by the camera to that presented by the lens. Multiply the weight of the lens by the distance of its cg from the center of the tripod and you have the torque presented by the lens. Add that to the torque presented by the camera and you have the total torque.
Topher925
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#3
Aug13-08, 10:25 AM
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P: 1,672
but this is more a function of torque than weight.
I don't mean to be picky but what you are after is the MOMENT not the torque. A torque is a dynamic torsional force, like found in the drive shaft of a car, the moment is a static torsion like what is represented here.

Anyway, like isly said just multiply the gravitational force of the lens and camera by the distance its offset from the hinge, assuming these vectors are orthonormal. If they are not, then just take the cross product.

ICLKennyg
ICLKennyg is offline
#4
Aug13-08, 10:35 AM
P: 2

Fulcrum problem (I think)


Quote Quote by isly ilwott View Post
The link wouldn't work for me.

The answer appears to be quite simple to figure though. If the cg (center of gravity) of the camera is exactly over the vertical shaft of the tripod, it will present 0 gram.centimeters torque to the head. If it is not, you must add (or subtract) the torque presented by the camera to that presented by the lens. Multiply the weight of the lens by the distance of its cg from the center of the tripod and you have the torque presented by the lens. Add that to the torque presented by the camera and you have the total torque.
Sorry misplaced my plurals
http://kj.stillabower.net/forums/img/camera.jpg

Do the units matter, or will they cancel as long as they are consistant?
isly ilwott
isly ilwott is offline
#5
Aug13-08, 10:44 AM
P: 72
Quote Quote by Topher925 View Post
I don't mean to be picky but what you are after is the MOMENT not the torque. A torque is a dynamic torsional force, like found in the drive shaft of a car, the moment is a static torsion like what is represented here.

Anyway, like isly said just multiply the gravitational force of the lens and camera by the distance its offset from hinge, assuming these vectors are orthonormal. If they are not, then just take the cross product.
That is picky. Should I rename my torque wrench to be a moment wrench?

All this time I've thought torque to be:

n.
The moment of a force; the measure of a force's tendency to produce torsion and rotation about an axis, equal to the vector product of the radius vector from the axis of rotation to the point of application of the force and the force vector.
A turning or twisting force

The object being torqued does not have to be rotating. The torsional force, quantified in foot-pounds, gram-centimeters, kilogram-meters or some other understandable units is still properly labeled as torque. My understanding is that structural engineers use the term "moment" to describe a similar static force on beams, that force being calculated much the same as torque on a hex head bolt is calculated...except it's more complicated than using one distance and one force.
isly ilwott
isly ilwott is offline
#6
Aug13-08, 10:47 AM
P: 72
Quote Quote by ICLKennyg View Post
Sorry misplaced my plurals
http://kj.stillabower.net/forums/img/camera.jpg

Do the units matter, or will they cancel as long as they are consistant?
They do not cancel. You will just get a different number if you calculate foot-pounds as opposed to gram-centimeters. You had originally expressed the linear measure in cm and the weight in grams.


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