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Category Ground based logging / Tractor winches
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TIGER DSU/WH 60E
General information
Technical data
Producer / distributor
Economy / calculation
Single drum winch for the three-point hitch with own oil supply, hydr. height-adjustable wire feed rollers, radio
Driving force kN:
60
Mass
kg:
550
Inner cable force
60 kN
Cable capacity
100
Diameter of cable
11 mm
Alternative cable
-
Avg. cable speed
-
Power transmission
of the tractor PTO via cardan shaft to the gear driving the winch
Gear
Oil bath - table drive
Clutch
plate clutch
Brake
band brake
Control
electro-hydraulic, radio-controlled
Optional equipment
-
Mass without cable
550 kg
Logging pan
-
Style
-
Number of drums
1
TIGER Seilwinden u. Maschinenbau GmbH&CoKG
Nussbacherstraße 23
A-4541 Adlwang
T: +43(0)7258/3901
E: office@tiger-seilwinden.at
I: www.tiger-seilwinden.at
TIGER Seilwinden u. Maschinenbau GmbH&CoKG
Nussbacherstraße 23
A-4541 Adlwang
T: +43(0)7258/3901
E: office@tiger-seilwinden.at
I: www.tiger-seilwinden.at
Type in your own numbers here and press
Annual workload
Annual workload
Number of operating hours in which a machine is used per year.
The annual workload cannot be higher than the maximum annual workload. If a bigger value is beeing put in, the maximum annual workload will be used by the program.
hrs.
BFW
KWF
FAO
Costs per hour
Costs per hour
In this calculation only direct costs for a machine/equipment are retrieved. This does not include labour costs for operation or transfer of machinery nor costs for administration, buildings etc. of an enterprise.
Depreciation
Cost of depreciation per operating hour = Pr / (Nr * JA)
Pr
...Price of machinery/equipment
Nr
...Reduced time of economic use in years
JA
...Annual workload in operating hours
Interest
Interest
Interest is calculated for one half of the price of the machinery for a reduced time of economic use. This ensures an interest which is the same for all years of use.
Expense of interest per operating hour = (Pr / 2) * (p / 100) * (Nr / Hr)
Pr
...Price
p
....Interest rate in percent
Nr
...Reduced time of use in years
Hr
...Reduced operating hours of use
Repair cost
Cost of repair
The calculation is done using a coefficient deduced from empiric values. This coefficient is the relation between cost of repair and the price of the new equipment/machinery for maximum operating hours of economic use. For a lower annual workload the cost of repair is reduced by a factor consisting of reduced hours of use divided by maximum hours of use. This takes the fact of a lower total amount of operating hours and therefore lower repair costs into consideration.
Cost of repair per operating hour = (Pr / H) * (Hr / H) * r
Pr
...Price of equipment/machinery
H
....Maximum hours of economic use
Hr
...Reduced duration of use in operating hours
r
....Cost of repair price coefficient
Garage cost
Cost of garage per operating hour = (V * KS) / JA
V
.....Required space in cubic meters
KS
...Cost of building per cubic meter and year in ¤
JA
...Annual workload in operating hours
Depreciation of tow rope
Suggested price
Suggested price
Suggested prices are equivalent to depreciation sums. Put in this price including expenses for parts like tires, cables or tools. For example for chainsaws prices are to be put in inclusive expenses for swivel blade and sawing chain.
For calculation the expenses for these parts are subtracted from the depreciation sum and calculated seperately. Reason for this are differences in the maximum number of hours of economic use.
In case of sled winches suggested price is given without cables, as cable equipment can vary very much in quantity and material. In this case you should put in your own price without cables.
¤
Max. operating hours of economic use
Max. operating hours of economic use
Total number of operating hours for a device at maximum annual workload and economic use.
hrs.
Max. years of economic use
Maximum economic useful life
Useful life in years, in which the machine can be operated with economic success.
The Maximum economic useful life is a hypothetical value which is only valid for an annual workload of null hours. Using an elliptic equation a reduced useful life ist derivated and used for further calculation.
years
Maximum annual workload
Maximum annual workload
Highest possible number of operating hours which one can use a machinery/equipment, independent from the situation of the own enterprise.
If this value is higher than the maximum operating hours of economic use then the maximum annual workload is set to the value for maximum operating hours of economic use, because our calculating routine cannot deal with a useful life lower than one year.
hrs.
Interest rate
Interest rate
Annual interest for the investition expressed in percent.
%
Cost of repair as a prize coefficient
Cost of repair as a price coefficient
The calculation is done using a coefficient deduced from empiric values. This coefficient is the relation between cost of repair and the price of the new equipment/machinery for maximum operating hours of economic use. For a lower annual workload the cost of repair is reduced by a factor consisting of reduced hours of use divided by maximum hours of use. This takes the fact of a lower total amount of operating hours and therefore lower repair costs into consideration.
Required garage space
Required garage space
Required space of machine/equipment in cubic meters.
If null, this number will be calculated if any values for length, width and height are found in the database.
m³
Cost of building per cubic meter/yr.
Cost of Building
Cost of Building per cubic meter and year in ¤.
¤/m³
Tow rope
Suggested price
¤
Max. operating hours of economic use
hrs.
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