You discovered our blog Arts And Crafts Ideas. Please Remember to bookmark this page Dimensional Finish. If you enjoy our post Dimensional Finish, show your love by hitting one of the social media buttons above for this page.
Dimensional Finish
Shimmerz and paperclay chipboard layout
Metal finishing technique adds to modernization and leasing Strategies
Technology is developing and progressing at quite a rapid pace today. It is not just adding more flexibility and comfort to our life but it is also making it fast and comfortable . Such new technology not just makes life easier for designers, contractors, building owners, leasing agents, tenants and facilities maintenance crews but for everyone else involved. There are so many techniques that are being introduced and invented from time to time to make human life more comfortable and reduce human effort. The technique of applied design is actually a process that allows the building owners to make a personalized artwork for metal panels, elevator hatchway doors and signage with various optical effects and colors. The applied designs can be changed overnight and as per your likings.
With newer designs one can get more options for decorating a particular area and enhance its overall ambience. A recent introduction of applied designed technique can be seen lobby of the four seasons hotel in New York. The doors of the hatchway are mirror finished around the elevator vestibule which is selectively retained to emphasize a symmetrical design. Afterwards oxidized and brushed satin finishes were used to outline the geometric shaped in its panels.
Afterwards the technique was chosen for a high profile job to impart a three dimensional, shimmering look to the design of the elevator's door. This technique helped in saving money and eliminated the need of new panels. The alteration in the design of the elevator panels cost 1/3 of the one half of the cost of new panels. Therefore, the technique of applied designs is not just able provide a new look to a particular area but also helps in cutting down the cost of change considerable thus adding to its utility and durability. Also this method can also be used to modernize the elevator as many times you want in stainless steel, clad in bronze or aluminum. Therefore, leasing agents and building owners can offer prospective, customized tenant doors with letterings, designs and logos for their dedicated elevator lines or floors.
The redesign consumes much lesser time than fitting new panels. The redesigning work can be done on the site within the same day saving time, effort and interruptions on the working days. It also eliminated the risk of transporting the panels for the contractors.
Applied redesign technology is different from the decorative etching which leaves metal surfaces un-refinishable. Redesign technology allows you to change, restore and refinish designs as many times as you like.
Therefore, you can help save a lot of money, time and resources of your company by choosing redesigning techniques over choosing new panels.
About the Author
Tom Kelly Website:www.indovance.com EMail:tkelly@indovance.com
If you are looking for a different item here are a list of related products on Arts And Crafts Ideas, please check out the following:
Frequently Asked Questions...
Physics help, one dimensional kinematics: track event?
During a track even two runners, Bob, and Jim, round the last turn and head into the final stretch with Bob 2.0 meters in front of Jim. They are both running with the same velocity 8.0 m/s. When the finish line is 48 meters away from Jim, Jim accelerates at 1.0 m/s^2 until he catches up to Bob. Jim then continues at a constant speed until he reaches the finish line.
a) How long did it take Jim to catch Bob?
b) How far did Jim still have to run when he just caught up to Bob?
c) How long did Jim take to reach the finish line after he just caught up to Bob?
Bob starts to accelerate when Jim just catches up to him, and accelerates all the way to the finish line and crosses the line exactly when Jim does. Assume Bob's acceleration is constant.
d) What is Bob's acceleration?
e) What is Bob's velocity at the finish line? Who is running faster?
Answer:
Let's try it the hard way, first.
Let's let t be time, and let it be zero when Jim starts to accelerate. We'll let s be position, and sJ and sB will be Bob's and Jim's position, respectively. Then, we'll let the finsh line be at s=0.
Jim's position is:
sJ = s0 + v0t + (at^2)/2,
where s0 = his initial position, v0 = his initial velocity, and a = his acceleration. From the problem, his initial position is 48 meters from the finish line, so s0 = -48 m. We know from the problem that v0 is 8 m/s, and a is 1.0 m/s^2, so Jim's positon is described by:
sJ = -48 + 8t + (t^2)/2
Bob's position is easier to calculate. It's the same as Jim's, but the acceleration is zero. Bob is 2 meters ahead of Jim, so his s0 is -46 instead of -48. So,
sB = -46 + 8t
Jim catches Bob when sJ = sB, or when
-48 + 8t + (t^2)/2 = -46 + 8t
The "8t" terms cancel, and adding 46 to each side of the equation yields
-2 + (t^2)/2 = 0, or (t^2)/2 = 2, or t^2 = 4
Obviously, t = 2. So that's the answer to a).
When Jim caught Bob, his position was
sJ = -48 + 8*2 + (2^2)/2, or
sJ = -48 + 16 + 2, reducing to 30 meters. That's b).
Jim didn't accelerate after he caught Bob. That was at t=2, so his velocity had increased by 2, from 8 m/s to 10. Let's reestablish t=0 when Jim passed Bob. Here's the new equation for his position:
sJ = -30 + 10t.
He reaches the finsh line when sJ = 0,
0 = 30 + 10t, or 30 = 10t, or t = 3. So, Jim crosses the line 3 seconds after he passes Bob. That's c).
Now, let's let Bob accelerate. He starts accelerating at t=0, and at that time, his position and Jim's are the same at -30. Jim's position is:
sJ = -30 + 10t
Bob's is sB = -30 + 8t + (at^2)/2
We know Jim crosses the finsh at t=3. If they cross at the same time, Bob's position is zero at t=3.
0 = -30 + 8*3 + (a*3^2)/2, and
0 = -30 + 24 + 9a/2
0 = -6 + 4.5a
6 = 4.5a, and a = 4/3 m/s^2. That's d).
Bob's velocity at the finish line is just his original velocity plus his acceleration multiplied by the amount of time he accelerated, or
vB = 8 + 3*4/3, or 12 m/s. That's part of e).
Who was going faster? Well, Bob was going 12 m/s, and Jim 10 m/s. Bob was going faster. That's the rest of e).
Now, let's look at it the easy way.
Jim and Bob start 2 meters apart. They're going at exactly equal velocities, so those don't matter. It's the same as if Jim and Bob were stopped, with Bob 2 meters ahead, and Jim starts to accelerate at 1 m/s^2 from a dead stop. When does he get to Bob?
0 = -2 + (at^2)/2 = -2 + t^2/2, and
t = 2. Same as before, a)
How far did he have to run after he caught Bob? Well, it took 2 seconds to catch Bob. Bob was 46 meters from the finish at t=0, and he went 8 m/s, so Bob was 30 meters away from the finish line at that time. Same as b).
How long did it take Jim to finsh after he caught up with Bob? He started at 8 m/s, and he accelerated at 1 m/s^2 for 2 seconds, so his velocity was 10 m/s. Same as c).
Working on d) and e), it's pretty much the same as before. But, obviously, we can eliminate some effort by looking only at the difference in Jim and Bob's frames of reference for the first three parts of the question.
